# Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.6 | Set 1

### Question 1. Find the sum of the following arithmetic progressions:

### (i) 50, 46, 42, â€¦ to 10 terms

**Solution: **

Given A.P. has first term(a) = 50,

Common difference(d) = 46 – 50 = – 4

and number of terms(n) = 10

So, Sum of A.P. = S

_{10}= n[2a + (n – 1)d] / 2= 10[2(50) + (10 – 1)(-4)]/2

= (10)(100 – 36)/2

= 320

Hence, the sum of the first 10 terms of A.P. is 320.

### (ii) 1, 3, 5, 7, â€¦..to 12 terms.

**Solution: **

Given A.P. has first term(a) = 1,

common difference(d) = 3 – 1 = 2

and number of terms(n) = 12

So, Sum of A.P. = S

_{12}= n[2a + (n – 1)d] / 2= 12[2(1) + (12 – 1)(2)]/2

= 12[2 + 22]/2

= 144

Hence, the sum of the first 12 terms of A.P. is 144.

### (iii) 3, 9/2, 6, 15/2, â€¦ to 25 terms

**Solution: **

Given A.P. has first term(a) = 3,

Common difference(d) = 9/2 – 3 = 3/2

and number of terms(n) = 25

So, Sum of A.P. = S

_{25}= n[2a + (n – 1)d] / 2= 25[2(3) + (25 – 1)(3/2)]/2

= 25[6 + 36]/2 = 525

Hence, the sum of the first 25 terms of A.P. is 525.

### (iv) 41, 36, 31, â€¦.. to 12 terms.

**Solution: **

Given A.P. has first term(a) = 41,

Common difference(d) = 36 – 41 = -5

and number of terms(n) = 12

So, Sum of A.P. = S

_{12}= n[2a + (n – 1)d] / 2= 12[2(41) + (12 – 1)(-5)]/2

= 12[82 – 55] = 162

Hence, the sum of the first 12 terms of A.P. is 162.

### (v) a + b, a â€“ b, a â€“ 3b, â€¦.. to 22 terms

**Solution:**

Given A.P. has first term(a) = a + b,

Common difference(d) = (a – b) – (a + b) = -2b

and number of terms(n) = 22

So, Sum of A.P. = S

_{22}= n[2a + (n – 1)d] / 2= 22[2(a + b) + (22 – 1)(-2b)]/2

= 11{2(a + b) – 22b)

= 11(2a – 40b) = 22a – 440b

Hence, the sum of the first 22 terms of A.P. is 22a â€“ 440b.

### (vi) (x â€“ y)^{2},(x^{2} + y^{2}), (x + y)^{2},â€¦ to n terms

**Solution:**

Given A.P. has first term(a) = (x – y)

^{2},number of terms(n) = n and

Common difference(d) = x

^{2}+ y^{2}– (x – y)^{2}= x^{2}+ y^{2}– x^{2}+ y^{2}+ 2xy = 2xySo, Sum of A.P. = S

_{n}= n[2a + (n – 1)d] / 2= n[2(x – y)

^{2}+(n – 1)(2xy)]/2= n[(x – y)

^{2}+(n – 1)(xy)]

Hence, the sum of the first n terms of A.P. is n[(x â€“ y)^{2}+(n â€“ 1)(xy)].

### (vii)

**Solution:**

Given A.P. has first term(a) =

number of terms(n) = n and

Common difference(d) =

=

So, Sum of A.P. = S

_{n}= n[2a + (n – 1)d] / 2=

=

=

Hence, the sum of the first n terms of A.P. is

### (viii) â€“26, â€“24, â€“22,…. to 36 terms

**Solution:**

Given A.P. has first term(a) = -26,

Common difference(d) = -24 – (-26) = 2

and number of terms(n) = n.

So, Sum of A.P. = S

_{36}= n[2a + (n – 1)d] / 2= 36[2(-26) + (36 – 1)2]/2

= 18[-52 + 70] = 324

Hence, the sum of the first 36 terms of A.P. is 324.

### Question 2. Find the sum to n terms of the A.P. 5, 2, â€“1, â€“ 4, â€“7, â€¦

**Solution:**

Given A.P. has first term(a) = 5,

Common difference(d) = 2 – 5 = -3

Sum of n terms of A.P. = S

_{n}= n[2a + (n – 1)d] / 2= n[2(5) + (n – 1)(-3)] / 2

= n{10 – 3n + 3)} / 2

= n / 2(13 – 3n)

Hence, the sum of the n terms of A.P. is n/2(13 – 3n).

### Question 3. Find the sum of n terms of an A.P. whose terms are given by a_{n} = 5 â€“ 6n.

**Solution:**

Given A.P. has nth term, an = 5 – 6n

Putting n = 1, we get a, first term of the A.P.,

a = 5 – 6(1) = -1

Sum of n terms of A.P. = S

_{n}= n[a + an]/2= n[-1 + (5 – 6n)]/2

= n(4 – 6n)/2 = n(2 – 3n)

Hence, the sum of the n terms of A.P. is n(2 â€“ 3n).

### Question 4. Find the sum of the last ten terms of the A.P. : 8, 10, 12, 14, .. , 126.

**Solution:**

Given A.P. has first term(a) = 8,

Common difference(d) = 10 – 8 = 2

and nth term(a

_{n}) = 126.The nth term of the A.P. is given by, a

_{n}= a+(n – 1)d=> 126 = 8 + (n – 1)(2)

=> 126 = 8 + 2n â€“ 2

=> 2n = 120

=> n = 60

We have to find the sum of last ten terms i.e.,

S = a

_{51}+ a_{52}+ a_{53}+ â€¦â€¦. + a_{60}a

_{51}= 8 + (51 – 1)(2) = 8 + 50(2) = 108So, sum would be S = 10[108 + 126]/2 = 5(234) = 1170

Hence, the sum of last 10 terms of the A.P. is 1170.

### Question 5. Find the sum of first 15 terms of each of the following sequences having nth term as:

### (i) a_{n} = 3 + 4n

**Solution:**

Given A.P. has nth term, a

_{n}= 3 + 4nand number of terms(n) = 15

So, a

_{n}= 3 + 4(15) = 63On putting n = 1, we get a, first term of the A.P.,

a = 3 + 4(1) = 7

S

_{15}= n[a + a_{n}]/2= 15(7 + 63)/2 = 15 x 35 = 525

Hence, the sum of the first 15 terms of given A.P. is 525.

### (ii) b_{n} = 5 + 2n

**Solution:**

Given A.P. has nth term, a

_{n}= 5 + 2n andnumber of terms(n) = 15

So, a

_{n}= 5 + 2(15) = 35On putting n = 1, we get a, first term of the A.P.,

a = 5 + 2(1) = 7

S

_{15}= n[a + a_{n}]/2= 15(7 + 35)/2 = 15 x 21 = 315

Hence, the sum of the first 15 terms of given A.P. is 315.

### (iii) x_{n} = 6 â€“ n

**Solution:**

Given A.P. has nth term, a

_{n}= 6 – nand number of terms(n) = 15

So, a

_{n}= 6 – n = -9On putting n = 1, we get a, first term of the A.P.,

a = 6 – 1 = 5

S

_{15}= n[a + a_{n}]/2= 15(5 – 9)/2 = 15 x (-2) = -30

Hence, the sum of the first 15 terms of given A.P. is -30.

**(iv) y _{n} = 9 â€“ 5n**

**Solution:**

Given A.P. has nth term, a

_{n}= 9 – 5nand number of terms(n) = 15

So, a

_{n}= 9 – 5(15) = -66On putting n = 1, we get a, first term of the A.P.,

a = 9 – 5 = 4

S

_{15}= n[a + a_{n}]/2= 15(4 – 66)/2 = 15 x (-31) = -465

Hence, the sum of the first 15 terms of given A.P. is -465.

### Question 6. Find the sum of first 20 terms the sequence whose nth term is a_{n} = An + B.

**Solution:**

Given A.P. has nth term, a

_{n}= An + Band number of terms(n) = 20

So, a

_{n}= A(20) + B = 20A + BOn putting n = 1, we get a, first term of the A.P.,

a = A(1) + B = A + B

S

_{20}= n[a + a_{n}]/2= 20(A+B + 20A + B)/2

= 10[21A + 2B]

= 210A + 20B

Hence, the sum of the first 20 terms of given A.P. is 210A + 20B.

### Question 7. Find the sum of first 25 terms of an A.P whose nth term is given by a_{n} = 2 â€“ 3n.

**Solution:**

Given A.P. has nth term, a

_{n}= 2 – 3nand number of terms(n) = 25

So, a

_{n}= 2 – 3(25) = -73On putting n = 1, we get a, first term of the A.P.,

a = 2 – 3(1) = -1

S

_{20}= n[a + a_{n}]/2= 25(-1 – 73)/2

= 25 x (-37) = -925

Hence, the sum of the first 25 terms of given A.P. is -925.

### Question 8. Find the sum of first 25 terms of an A.P whose nth term is given by a_{n} = 7 â€“ 3n.

**Solution:**

Given A.P. has nth term, a

_{n}= 7 – 3nand number of terms(n) = 25

So, a

_{n}= 7 – 3(25) = -68On putting n = 1, we get a, first term of the A.P.,

a = 7 – 3(1) = 4

S

_{25}= n[a + a_{n}]/2= 25(4 – 68)/2

= 25 x (-32)

= -800

Hence, the sum of the first 25 terms of given A.P. is -800.

### Question 9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.

**Solution:**

Given A.P. has first term(a) = 25,

Common difference(d) = 22 – 25 = -3 and sum(S

_{n}) = 116.We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n – 1)d] / 2.=> 116 = n[2(25) + (n âˆ’ 1)(âˆ’3)]/2

=> n[53 âˆ’ 3n]/2 = 116

=> 53n – 3n

^{2}= 232=> 3n

^{2}– 53n + 232 = 0=> 3n

^{2}â€“ 24n – 29n + 232 = 0=> 3n(n – 8) â€“ 29 (n – 8) = 0

=> (3n – 29)( n – 8 ) = 0

=> n = 29/3 or n = 8

Ignoring n = 29/3 as number of terms cannot be a fraction, so we get, n = 8.

So, the last term is:

a

_{8}= a + (8 – 1)d = 25 + 7(-3) = 25 – 21 = 4

Hence, the last term of the given A.P. is 4.

### Question 10 (i). How many terms of the sequence 18, 16, 14â€¦. should be taken so that their sum is zero.

**Solution:**

Given A.P. has first term(a) = 18,

Common difference(d) = 16 – 18 = -2 and sum(S

_{n}) = 0.We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n – 1)d] / 2.=> 0 = n[2(18) + (n – 1)(-2)]/2

=> 0 = n[36 + (-2n + 2)]/2

=> n[38 âˆ’ 2n] = 0

=> n = 0 or 38 – 2n = 0

Ignoring n = 0 as number of terms cannot be 0. So we get,

=> 38 – 2n = 0

=> 2n = 38

=> n = 19

Hence, the number of terms(n) is 19.

### (ii) How many terms are there in the A.P. whose first and fifth terms are â€“14 and 2 respectively and the sum of the terms is 40?

**Solution:**

Given A.P. has first term(a) = â€“14,

Fifth term(a

_{5}) = 2 and sum(S_{n}) = 40.Let the common difference of the A.P. be d. We get,

Then, a

_{5}= a + (5 – 1)d=> 2 = -14 + 4d

=> 4d = 16

=> d = 4

We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n – 1)d] / 2.=> 40 = n[2(-14) + (n – 1)(4)]/2

=> 40 = n[-28 + (4n – 4)]/2

=> 40 = n[-32 + 4n]/2

=> 4n

^{2}– 32n – 80 = 0=> n

^{2 }– 8n – 20 = 0=> n

^{2}-10n + 2n – 20 = 0=> n(n -10) + 2(n – 10 ) = 0

=> (n + 2)(n – 10) = 0

=> n = -2 or n = 10

Ignoring n = -2 as number of terms cannot be negative. So we get, n = 10.

Hence, the number of terms (n) is 10.

**(**iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636?

**Solution:**

Given A.P. has first term(a) = 9,

Common difference(d) = 17 – 9 = 8 and sum(S

_{n}) = 636.We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n – 1)d] / 2.=> 636 = n[2(9) + (n âˆ’ 1)(8)]/2

=> 636 = n[18 + (8n âˆ’ 8)]/2

=>1271 = 10n + 8n

^{2}=> 8n

^{2}+ 10n âˆ’ 1272 = 0=> 4n

^{2}+ 5n âˆ’ 636 = 0=> 4n

^{2}âˆ’ 48n + 53n âˆ’ 636 = 0=> 4n(n âˆ’ 12) + 53(n âˆ’ 12) = 0

=> (4n + 53)(n âˆ’ 12) = 0

=> n = âˆ’53/4 or n = 12

Ignoring n = âˆ’53/4 as number of terms cannot be fraction. So we get, n = 12.

Hence, the number of terms (n) is 12.

### (iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693?

**Solution:**

Given A.P. has first term(a) = 63,

Common difference(d) = 60 – 63 = -3 and sum(S

_{n}) = 636.We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n – 1)d] / 2.=> 693 = n[2(63) + (n âˆ’ 1)(âˆ’3)]/2

=> 693 = n[126+(âˆ’3n + 3)]/2

=> 693 = n[129 âˆ’ 3n]/2

=> 129n âˆ’ 3n

^{2}= 1386=> 3n

^{2}âˆ’ 129n + 1386 = 0=> n

^{2}âˆ’ 43n + 462 = 0=> n

^{2}âˆ’ 22n âˆ’ 21n + 462 = 0=> n(n âˆ’ 22) âˆ’ 21(n âˆ’ 22) = 0

=> (n âˆ’22) (n âˆ’21) = 0

=> n = 22 or n = 21

So, the value of n can be both 21 as well as 22 because,

a

_{22}= a + 21d = 63 + 21(âˆ’ 3) = 63 âˆ’ 63 = 0Sum remains the same even if we add the 22nd term because its value is 0.

Hence, the number of terms (n) is 21 or 22.

### (v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero?

**Solution:**

Given A.P. has first term(a) = 27,

common difference(d) = 24 âˆ’ 27 = âˆ’3

and sum(S

_{n}) = 0.We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.=> 0 = n[2(27) + (n âˆ’ 1)( âˆ’ 3)]/2

=> 0 = n[54 + (n âˆ’ 1)(-3)]

=> 0 = n[54 âˆ’ 3n + 3]

=> n[57 âˆ’ 3n] = 0

=> n = 0 or 3n = 57

Ignoring n = 0 as number of terms cannot be zero. So we get,

=> 3n = 57

=> n = 19

Hence, the number of terms (n) is 19.

### Question 11. Find the sum of the first

### (i) 11 terms of the A.P. : 2, 6, 10, 14, . . .

**Solution:**

Given A.P. has first term(a) = 2,

Common difference(d) = 6 âˆ’ 2 = 4

and number of terms(n) = 11.

We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{11}= 11[2(2) + (11 âˆ’ 1)4]/2= 11[4 + 40]/2

= 11 Ã— 22 = 242

Hence, the sum of first 11 terms of the given A.P. is 242.

### (ii) 13 terms of the A.P. : -6, 0, 6, 12, . . .

**Solution:**

Given A.P. has first term(a) = -6,

Common difference(d) = 0 âˆ’ (âˆ’6) = 6

and number of terms(n) = 13.

We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{13}= 13[2(âˆ’6) + (13 âˆ’ 1)6]/2= 13[(âˆ’12) + 72]/2

= 13[30] = 390

Hence, the sum of first 13 terms of the given A.P. is 390.

### (iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.

**Solution:**

Given A.P. has second term(a

_{2}) = 2,Fourth term(a

_{4}) = 8 andnumber of terms(n) = 51.

=> a

_{2}= a + d=> 2 = a + d â€¦(1)

Also, a

_{4}= a + 3d=> 8 = a + 3d â€¦ (2)

Subtracting (1) from (2), we have

=> 2d = 6

=> d = 3

Putting d = 3 in (1), we get a = âˆ’1.

We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{51}= 51[2(âˆ’1) + (51 âˆ’ 1)(3)]/2= 51[âˆ’2 + 150]/2

= 51[74] = 3774

Hence, the sum of first 51 terms of the given A.P. is 3774.

### Question 12. Find the sum of

### (i) the first 15 multiples of 8

**Solution:**

First 15 multiples of 8 are 8, 16, 24, â€¦â€¦ , 120.

These multiples form an A.P. with first term(a) = 8,

Common difference(d) = 16 âˆ’ 8 = 8

and number of terms(n) = 15.

We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{15}= 15[2(8) + (15 âˆ’ 1)8]/2= 15[16 + 112]/2

= 15[64] = 960

Hence, the sum of the first 15 multiples of 8 is 960.

### (ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

**Solution: **

We know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.

(a)First 40 positive integers divisible by 3 are 3, 6, 9, 12,â€¦â€¦ ,120.These integers form an A.P. with first term(a) = 3,

Common difference(d) = 6 âˆ’ 3 = 3 and number of terms(n) = 40.

S

_{40}= 40[2(3) + (40 âˆ’ 1)3]/2= 40(6 + 117)/2

= 20(123) = 2460

Hence, the sum of first 40 multiples of 3 is 2460.

(b)First 40 positive integers divisible by 5 are 5, 10, 15, 20,â€¦â€¦ ,200.These integers form an A.P. with first term(a) = 5,

Common difference(d) = 10 âˆ’ 5 = 5 and number of terms(n) = 40.

S

_{40}= 40[2(5) + (40 âˆ’ 1)5]/2= 40(10 + 195)/2

= 20(205) = 4100

Hence, the sum of first 40 multiples of 5 is 4100.

(c)First 40 positive integers divisible by 6 are 6, 12, 18, 24,â€¦â€¦ ,240.These integers form an A.P. with first term(a) = 6,

Common difference(d) = 12 âˆ’ 6 = 6 and number of terms(n) = 40.

S

_{40}= 40[2(6) + (40 âˆ’ 1)6]/2= 40(12 + 234)/2

= 20(246) = 4920

Hence, the sum of first 40 multiples of 6 is 4920.

### (iii) all 3â€“digit natural numbers which are divisible by 13.

**Solution:**

All 3â€“digit natural numbers which are divisible by 13 are 104, 117,â€¦â€¦ ,988.

These numbers form an A.P. with first term(a) = 104 and

Common difference(d) = 117 âˆ’ 104 = 13.

We know, the nth term of an A.P. id given by, a

_{n}= a + (n âˆ’ 1)d.=> 988 = 104 + (n âˆ’ 1)13

=> 988 = 104 + 13n -13

=> 988 = 91 + 13n

=> 13n = 897

=> n = 69

Also, we know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{69}= 69[2(104) + (69 âˆ’ 1)13]/2= 69[1092]/2

= 69(546) = 37674

Hence, the sum of all 3â€“digit natural numbers which are divisible by 13 is 37674.

### (iv) all 3â€“digit natural numbers which are multiples of 11.

**Solution:**

All 3â€“digit natural numbers which are divisible by 11 are 110, 121, 132,â€¦â€¦ ,990.

These numbers form an A.P. with first term(a) = 110 and

Common difference(d) = 121 âˆ’ 110 = 11.

We know, the nth term of an A.P. id given by, a

_{n}= a + (n âˆ’ 1)d.=> 990 = 110 + (n âˆ’ 1)11

=> 990 = 110 + 11n -11

=> 990 = 99 + 11n

=> 11n = 891

=> n = 81

Also, we know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{81}= 81[2(110) + (81 âˆ’ 1)11]/2= 81[1100]/2

= 81(550) = 44550

Hence, the sum of all 3â€“digit natural numbers which are divisible by 13 is 44550.

### (v) all 2-digit natural numbers divisible by 4.

**Solution:**

All 2-digit natural numbers divisible by 4 are 12, 16, 20,â€¦â€¦ ,96.

These numbers form an A.P. with first term(a) = 4

and common difference(d) = 16 âˆ’ 12 = 4.

We know, the nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 96 = 12 + (n âˆ’ 1)4

=> 4(n âˆ’ 1) = 84

=> n âˆ’ 1 = 21

=> n = 22

Also, we know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n âˆ’ 1)d] / 2.S

_{22}= 22[2(12) + (22 âˆ’ 1)4]/2= 22[24 + 84]/2

= 22[54] = 1188

Hence, the sum of all 2-digit natural numbers divisible by 4 is 1188.

### Question 13. Find the sum:

### (i) 2 + 4 + 6 + . . . + 200

**Solution:**

Given series is an A.P. with first term(a) = 2,

Common difference(d) = 4 âˆ’ 2 = 2 and nth term(a

_{n}) = 200.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 200 = 2 + (n âˆ’ 1)2

=> 200 = 2 + 2n âˆ’ 2

=> n = 200/2

=> n = 100

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{100 }= 100[2 + 200]/2= 100[101] = 10100

Hence, the sum of terms of the given series is 10100.

### (ii) 3 + 11 + 19 + . . . + 803

**Solution:**

Given series is an A.P. with first term(a) = 3,

Common difference(d) = 11 âˆ’ 3 = 8 and nth term(a

_{n}) = 803.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 803 = 3 + (n âˆ’ 1)8

=> 803 = 3 + 8n âˆ’ 8

=> n = 808/8

=> n = 101

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{101 }= 101[3 + 803]/2= 101[403]

= 40703

Hence, the sum of terms of the given series is 40703.

### (iii) (-5) + (-8) + (-11) + . . . + (-230)

**Solution:**

Given series is an A.P. with first term(a) = -5,

Common difference(d) = (âˆ’8) âˆ’ (âˆ’5) = âˆ’3 and nth term(a

_{n}) = âˆ’230.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> âˆ’230 = âˆ’5 + (n âˆ’ 1)(âˆ’3)

=> âˆ’230 = âˆ’5 âˆ’ 3n + 3

=> n = 228/3

=> n = 76

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{76}= 76 [âˆ’5 + (âˆ’230)]/2= 38 x (âˆ’235) = âˆ’8930

Hence, the sum of terms of the given series is â€“8930.

### (iv) 1 + 3 + 5 + 7 + . . . + 199

**Solution:**

Given series is an A.P. with first term(a) = 1,

Common difference(d) = 3 âˆ’ 1 = 2 and nth term(a

_{n}) = 199.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 199 = 1 + (n âˆ’ 1)(2)

=> 199 = 1 + 2n âˆ’ 2

=> n = 200/2

=> n = 100

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{100}= 100[1 + 199]/2= 50(200) = 10000

Hence, the sum of terms of the given series is 10000.

### (v) 7 + 10^{1}/_{2} + 14 + . . . + 84

**Solution:**

Given series is an A.P. with first term(a) = 7,

Common difference(d) = 10

^{1}/_{2}âˆ’ 7 = (21 âˆ’ 14)/2 = 7/2 and nth term(a_{n}) = 84.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 84 = 7 + (n âˆ’ 1)(7/2)

=> 168 = 14 + 7n âˆ’ 7

=> n = 161/7

=> n = 23

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{23}= 23[7 + 84]/2= 23(91)/2

= 2093/2 = 1046.5

Hence, the sum of terms of the given series is 1046.5

### (vi) 34 + 32 + 30 + . . . + 10

**Solution:**

Given series is an A.P. with first term(a) = 34,

Common difference(d) = 32 âˆ’ 34 = âˆ’2 and nth term(a

_{n}) = 10.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 10 = 34 + (n âˆ’ 1)(âˆ’2)

=> 10 = 34 âˆ’ 2n + 2

=> n = (36 âˆ’10)/2

=> n = 13

Also, we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{13}= 13[34 + 10]/2= 13(22)

= 286

Hence, the sum of terms of the given series is 286.

**(vii) 25 + 28 + 31 + . . . + 100 **

**Solution:**

Given series is an A.P. with first term(a) = 25,

Common difference(d) = 28 âˆ’ 25 = 3 and nth term(a

_{n}) = 100.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 100 = 25 + (n âˆ’ 1)(3)

=> 100 = 25 + 3n âˆ’ 3

=> n = 88/3

=> n = 26

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{26}= 26[25 + 100]/2= 13(125)

= 1625

Hence, the sum of terms of the given series is 1625.

### (viii) 18 + 15^{1}/_{2} + 13 + . . . + (â€“49^{1}/_{2})

**Solution:**

Given series is an A.P. with first term(a) = 18,

Common difference(d) = âˆ’ 18 = (31 âˆ’ 36)/2 = âˆ’5/2 and nth term(a

_{n}) =We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=>

_{ }= 18+(n âˆ’ 1)(âˆ’5/2)=> â€“135/2 = (n âˆ’ 1)(âˆ’5/2)

=> n âˆ’ 1 = 135/5

=> n = 28

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{28}= 28[18 + (âˆ’49^{1}/_{2})]/2= 14[âˆ’63/2]

= âˆ’441

Hence, the sum of terms of the given series isâˆ’441.

### Question 14. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

**Solution:**

Given A.P. has first term(a) = 17,

Common difference(d) = 9 and last term(a

_{n}) = 350.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d.=> 350 = 17 + (n âˆ’ 1) 9

=> 350 = 17 + 9n âˆ’ 9

=> 9n = 342

=> n = 38

Also we know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.S

_{38}= 38(17 + 350)/2= 19(367) = 6973

Hence, the number of terms of the given A.P is 38 and sum is 6973.

### Question 15. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

**Solution:**

Given A.P. has third term(a

_{3}) = 7 and seventh term(a_{7}) = 3a_{3}+ 2 = 3(7) + 2 = 23.We know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d. So, we get,a + 2d = 7 â€¦.. (1)

a + 6d = 23 â€¦.. (2)

Subtracting (1) from (2), we get,

=> (a + 6d) âˆ’ (a + 2d) = 23 âˆ’ 7

=> 4d = 16

=> d = 4

On putting d = 4 in (1), we get,

=> a + 2(4) = 7

=> a = 7 âˆ’ 8

=> a = âˆ’1

We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n âˆ’ 1)d] / 2.Here a = âˆ’1, d = âˆ’4, n = 20. So sum is,

S

_{20}= 20[2(âˆ’1) + (20 âˆ’ 1)(4)]/2= 20[-2 + 76]/2

= 20[39] = 740

Hence, the sum of first 20 terms for the given A.P. is 740.

### Question 16. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

**Solution:**

Given A.P. has first term(a) = 2, last term(a

_{n}) = 50 and sum(S_{n}) = 442.We know sum of n terms of an A.P. is given by S

_{n}= n[a + a_{n}]/2.=> 442 = n[2 + 50]/2

=> 26n = 442

=> n = 17

Also, we know nth term of an A.P. is given by, a

_{n}= a + (n âˆ’ 1)d. So, we get,=> 50 = 2 + (17 âˆ’ 1)d

=> 16d = 48

=> d = 3

Hence, the common difference of the A.P. is 3.

### Question 17. If 12th term of an A.P. is â€“13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

**Solution:**

We are given,

12th term of the A.P., (a

_{12}) = a + 11d = âˆ’13 â€¦. (1)Sum of first four terms = S

_{4}= 4[2a + (4 âˆ’ 1)d]/2 = 24=> 24 = 4[2a + 3d]/2

=> 2a + 3d = 12 â€¦.. (2)

On multiplying eq(1) by 2 and subtracting eq(2) from it we get,

=> (2a+3d) âˆ’ 2(a+11d) = 12 âˆ’ 2(âˆ’13)

=> 2a + 3d âˆ’ 2a âˆ’ 22d = 38

=> âˆ’19d = 38

=> d = âˆ’2

On putting the value of d in eq(1), we get,

=> a + 11(âˆ’2) = âˆ’13

=> a = âˆ’13 + 22

=> a = 9

Now, sum of first 10 terms is given by,

S

_{10}= 10[2(9) + (10 âˆ’ 1)(âˆ’2)]/2= 10(18 âˆ’ 18)/2 = 0

Hence, the sum of first 10 terms of the given A.P. is 0.

### Question 18. Find the sum of n terms of the series

**Solution:**

Given A.P. has first term(a) =

Common difference(d) =

=

= -1/n

We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n âˆ’ 1)d] / 2=

=

=

=

Hence, the sum of n terms of the given A.P. is

### Question 19. In an A.P., if the first term is 22, the common difference is â€“4 and the sum to n terms is 64, find n.

**Solution:**

Given A.P. has first term(a) = 22, common difference(d) = âˆ’4 and sum(S

_{n}) = 64.We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n âˆ’ 1)d] / 2.=> 64 = n[2(22) + (n âˆ’ 1)(âˆ’4)]/2

=> 64 = n[44 âˆ’ 4n + 4]/2

=> 48n âˆ’ 4n

^{2 }=128=> 4n

^{2}âˆ’ 48n + 128 = 0=> n

^{2}âˆ’ 12n + 32 = 0=> n

^{2}âˆ’ 8n âˆ’ 4n + 32 = 0=> n(n âˆ’ 8) âˆ’ 4(n âˆ’ 8) = 0

=> (n âˆ’ 8)(n âˆ’ 4) = 0

=> n = 8 or n = 4

Hence, the number of terms are either 4 or 8.

### Question 20. In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?

**Solution:**

Given A.P. has,

Fifth term, a

_{5}= a + 4d = 30 â€¦..(1)Twelfth term, a

_{12}= a + 11d = 65 â€¦.(2)On subtracting eq(1) from (2), we get,

=> (a + 11d) âˆ’ (a + 4d) = 65 âˆ’ 30

=> 7d = 35

=> d = 5

On putting d = 5 in eq(1), we get,

=> a + 4(5) = 30

=> a = 30 âˆ’ 20

=> a = 10

We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n âˆ’ 1)d] / 2.Here a = 10, d = 5 and n = 20. So we get,

S

_{20}= 20[2(10) + (20 âˆ’ 1)(5)]/2= 20[20 + 95]/2

= 10[115] = 1150

Hence, the sum of first 20 terms for the given A.P. is 1150.

### Question 21. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

**Solution:**

Given A.P. has,

Second term, a

_{2}= a + d = 14 â€¦..(1)Third term, a

_{3 }= a + 2d = 18Hence, common difference(d) = a

_{3}âˆ’ a_{2}= 18 âˆ’ 14 = 4.On putting d = 4 in (1), we get,

=> a + 4 = 14

=> a = 10

We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n âˆ’ 1)d] / 2.Here a = 10, d = 4 and n = 51. So we get,

S

_{51}= 51[2(10) + (51 âˆ’ 1)(4)]/2= 51[20 + 200]/2

= 51[110] = 5610

Hence, the sum of first 51 terms for the given A.P. is 5610.

### Question 22. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

**Solution:**

We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n âˆ’ 1)d]/2.So we get, S

_{7}= 49=> 7[2a + (7 âˆ’ 1)d]/2 = 49

=> 7[a + 3d] = 49

=> a + 3d = 7 â€¦.. (1)

And also, S

_{17}= 289=> 17[2a + (17 âˆ’ 1)d]/2 = 289

=> 17[a + 8d] = 289

=> a + 8d = 17 â€¦.. (2)

On subtracting eq(1) from (2), we get,

=> a + 8d âˆ’ (a + 3d) = 17 âˆ’ 7

=> 5d = 10

=> d = 2

On putting d = 2 in (1), we get,

=> a + 3(2) = 7

=> a = 1

Here a = 1, d = 2, so sum of n terms would be,

S

_{n}= n[2(1) + (n âˆ’ 1)(2)]/2= n[2 + 2n âˆ’ 2]/2

= n[n] = n

^{2}

Hence, the sum of n terms of the given A.P is n^{2}.

### Question 23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

**Solution:**

Given A.P. has first term(a) = 5, last term(a

_{n}) = 45 and sum(S_{n}) = 400.We know sum of n terms of an A.P. is given by, S

_{n}= n[a + a_{n}]/2=> 400 = n[5 + 45]/2

=> 50n = 800

=> n = 16

Also, we know nth term of an A.P. given by, a

_{n}= a + (n âˆ’ 1)d.=> 45 = 5 + (16 âˆ’ 1)d

=> 15d = 40

=> d = 8/3

Hence, the number of terms of given A.P. is 16 and common difference is 8/3.

### Question 24. In an A.P. the first term is 8, nth term is 33 and the sum of first n terms is 123. Find n and the d, the common difference.

**Solution:**

Given A.P. has first term(a) = 8, nth term(a

_{n}) = 33 and sum(S_{n}) = 123.We know sum of n terms of an A.P. is given by, S

_{n}= n[a + a_{n}]/2=> 123 = n[8 + 33]/2

=> 41n = 246

=> n = 6

Also, we know nth term of an A.P. given by, a

_{n}= a + (n âˆ’ 1)d.=> 33 = 8 + (6 âˆ’ 1)d

=> 5d = 25

=> d = 5

Hence, the number of terms of given A.P. is 6 and common difference is 5.

## Please

Loginto comment...