# Class 11 RD Sharma Solutions- Chapter 19 Arithmetic Progressions- Exercise 19.2 | Set 2

General Formula

Tn = a + (n – 1) * dwhere,

Tnis the nth term

ais the first term

d= common difference

Nth term form end

Tn = l – (n – 1) * dwhere,

Tnis the nth term

lis the first term

d= common difference

**Question 13. If (m + 1)**^{th} term of an A.P. is twice the (n + 1)^{th} term, Prove that (3m + 1)^{th} term is twice the (m+ n + 1)^{th} term.

^{th}term of an A.P. is twice the (n + 1)

^{th}term, Prove that (3m + 1)

^{th}term is twice the (m+ n + 1)

^{th}term.

**Solution:**

We are given, a

_{m+1}= 2 a_{n+1}To prove, a

_{3m+1}= 2a_{m+n+1}Lets suppose we have first term as a and common difference as d, then we are given

a + md = 2(a + nd)

a + md = 2a + 2nd

a = md- 2nd —–1

Now LHS = a

_{3m+1}= a + 3md

= md – 2nd +3md putting value of a

= 4md – 2nd

Now RHS = 2a

_{m+n+1}= 2( a + (m+n)d)

= 2( a +md +nd)

= 2 (md-2nd+md+nd) putting value of a

= 2(2md – nd)

= 4md – 2nd

So we can see that LHS = RHS.

**Question 14. If the n**^{th} term of the A.P. 9, 7, 5, … is same as the n^{th} term of the A.P. 15, 12, 9, … Find N.

^{th}term of the A.P. 9, 7, 5, … is same as the n

^{th}term of the A.P. 15, 12, 9, … Find N.

**Solution:**

We are given two AP. For first AP

a

_{1}= 9d

_{1}= 7 – 9 = -2T

_{1n}= a_{1}+ (n – 1)d_{1}= 9 + (n – 1)(-2)

= 11 – 2n

For second AP,

a

_{2}= 15d

_{2 }= 12 – 15 = -3T

_{2n}= a_{2}+ (n – 1)d_{2}= 15 +(n-1)(-3)

= 18 – 3n

According to question,

T

_{1n}= T_{2n}11 – 2n = 18 – 3n

n = 7

Hence 7

^{th}term is same in both AP.

**Question 15. i) Find the 12th term from the end on the following Arithmetic Progression: 3, 5, 7, 9,… 201.**

**Solution:**

We are given, first term, a = 3

last term, l = 201

common difference, d = 5 – 3 = 2

n = 12

So, the 12th term from last is , = 201 – (12 -1)*2

= 201 – 22

= 179

Hence, the 12th term from the end is 179.

**ii) Find the 12th term from the end of the following Arithmetic Progression: 3, 8, 13,… 253.**

**Solution:**

We are given, first term, a = 3

last term, l = 253

common difference, d = 8 – 3 = 5

n = 12

So, the 12th term from last is, = 253 – (12 -1)*5

= 253 – 55

= 198

Hence, the 12th term from the end is 198.

**iii) Find the 12th term from the end of the following Arithmetic Progression: 1, 4, 7, 10,… 88.**

**Solution:**

We are given, first term, a = 1

Last term, l = 88

common difference, d = 4 – 1 = 3

n = 12

So, the 12th term from last is, = 88 – (12 -1)*3

= 88 – 33

= 55

Hence, the 12th term from the end is 55.

**Question 16. The 4th Term of an A.P. is Three Times the First and the 7th Term Exceeds Twice the Third Term by 1. Find the First Term and the Common Difference.**

**Solution:**

Let’s suppose first term is a and common difference is d

We are given, a

_{4}= 3a_{1}a

_{7}= 2a_{3}+1So ,

a + 3d = 3a

2a – 3d = 0 —–1

And a + 6d = 2(a+2d) +1

a + 6d = 2a + 4d +1

a -2d = -1 —–2

On solving 1 and 2

2a – 3d = 0

-2a + 4d = 2d = 2

Putting value of d in equation 1

a = (3*d)/2

a = 6/2 = 3

Hence the first term is 3 and common difference is 2.

**Question 17. Find the Second Term and Nth Term of an A.P. Whose 6th Term is 12 and the 8th Term is 22.**

**Solution:**

Let’s suppose first term is a and common difference is d

We are given, a

_{6}= 12a

_{8}= 22So, a + 5d =12 —–1

a + 7d = 22 —–2

On solving 1 and 2

a + 7d – (a + 5d )= 22 – 12

7d – 5d = 10

d = 5

Putting value of d in equation 1

a + 25 = 12

a = 12- 25

a = -13

So second term is, a + d = -13 + 5 = -8

And n

^{th}term is T_{n}= -13 + (n-1)*5= 5n – 18

**Question 18. How Many Numbers of **Two-Digit** Are Divisible by 3?**

**Solution:**

First two digit number that is divisible bye 3 is 12. So a =3

Next two digit number that is divisible bye 3 is 15. So common difference is 15 – 12 = 3

Last two digit number that is divisible bye 3 is 99.

So we can apply general formula,

3 + (n-1)*3 = 99

(n-1)*3 = 96

n -1 = 32

n =33

Hence, 33 two digit numbers are divisible by 3.

**Question 19. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, Find 32nd term.**

**Solution:**

We are given, a = 7

l = 125

total terms = 60

If we have total 60 terms, then last term would be a

_{60}a + 59d =125

Putting value of a,

7 + 59d = 125

59d = 118

d= 2

So, 32nd term a

_{32 }= a + 31d= 7 + 31*2

= 7 + 62

= 69

Hence, 32nd term is 69.

**Question 20. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the First Term and the Common Difference of the A.P.**

**Solution:**

Let’s suppose first term is a and common difference is d. We are given,

a

_{4}+ a_{8}= 24a

_{6 }+ a_{10}= 34a+ 3d + a +7d = 24

2a + 10d = 24

a + 5d = 12 —–1

a + 5d + a+9d = 34

2a + 14d = 34

a + 7d = 17 ——2

On solving 1 & 2 we get,

a + 7d – (a + 5d) = 17 – 12

a + 7d – a – 5d = 5

2d = 5

d = 5/2

Putting value of d in equation 1

a + 5*(5/2) = 12

a = 12 – 25/2

a = (24- 25) /2

a = -1/2

Hence the first term is -1/2 and common difference is 5/2.

**Question 21. How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4?**

**Solution:**

Let’s make an AP of those numbers which when divided by 7 leaves remainder 4.

First number which when divided by 7 leaves remainder 4 is 4. So a

_{1}= 4Next number which when divided by 7 leaves remainder 4 is 11. So a

_{2}= 11Largest 3 digit number which when divided by 7 leaves remainder 4 is 998. So a

_{n}= 998So our AP is 4, 11, …. 998 and common difference is 11-4 = 7

So,

4 + (n – 1)*7 = 998

(n – 1)*7 = 994

n – 1 = 142

n =143

Hence, 143 numbers are which when divided by 7 leave remainder 4.

**Question 22. The first and the last terms of an A.P. are a and l respectively. Show that the sum of nth term from the beginning and nth term from the end is a + l.**

**Solution:**

Let’s suppose common difference is d.

We know that nth term from beginning is a + (n-1)d

And nth term from end is l – (n – 1)*d

To prove: nth term from beginning + nth term from end = a + l

LHS= nth term from beginning + nth term from end

= a + (n-1)d + l – (n – 1)*d

= a +l

= RHS

Hence proved.

**Question 23. If an A.P. is such that a**_{4}/a_{7} = 2/3, Find a_{6}/a_{8}.

_{4}/a

_{7}= 2/3, Find a

_{6}/a

_{8}.

**Solution:**

Let’s suppose first term is a and common difference is d.

We are given, (a + 3d)/(a+6d) = 2/3

3(a + 3d) =2(a+6d)

3a + 9d = 2a + 12d

a = 3d

Now, we have to find a

_{6}/a_{8}= (a + 5d)/ (a+7d)

Putting value of a

= (3d + 5d)/(3a + 7d)

= 8d/10d

= 4/5

Hence a

_{6}/a_{8}is 4/5.