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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 21 Some Special Series- Exercise 21.2

### Question 1. 3 + 5 + 9 + 15 + 23 + . . . . n terms

Solution:

We are given the series: 3 + 5 + 9 + 15 + 23 + . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 3 + 5 + 9 + 15 + 23 + . . . . + an–1 + an  . . . .(1)

S = 3 + 5 + 9 + 15 + 23 + . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [3 + (5 + 9 + 15 + 23 + . . . . + an–1 + an)] – [(3 + 5 + 9 + 15 + 23 + . . . . + an–2 + an–1) +an]

=> 0 = 3 + [(5 – 3) + (9 – 5) + (15 – 9) + . . . . + (an – an–1)] – an

=> an = 3 + [2 + 4 + 6 + . . . . (n–1) terms]

As the series 2+ 4 + 6 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 4–2 = 2. So, we get,

=> an = 3 + (n–1) [2(2)+(n–2)2]/2

= 3+ (n–1) [4+2n–4]/2

= 3 + n(n–1)

= n2 – n+3

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 2. 2 + 5 + 10 + 17 + 26 + . . . . n terms

Solution:

We are given the series: 2 + 5 + 10 + 17 + 26 + . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 2 + 5 + 10 + 17 + 26 + . . . . + an–1 + an  . . . .(1)

S = 2 + 5 + 10 + 17 + 26 + . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [2 + (5 + 10 + 17 + 26 + . . . . + an–1 + an)] – [(2 + 5 + 10 + 17 + 26 + . . . . + an–2 + an–1) +an]

=> 0 = 2 + [(5 – 2) + (10 – 5) + (17 – 10) + . . . . + (an – an–1)] – an

=> an = 2 + [3 + 5 + 7 + . . . . (n–1) terms]

As the series 3 + 5 + 7 + . . . . (n–1) terms is an A.P., with first term(a) = 3 and common difference(d) = 5–3 = 2. So, we get,

=> an = 2 + (n–1) [2(3)+(n–2)2]/2

= 2+ (n–1) [6+2n–4]/2

= 2 + (n–1)(n+1)

= n2 – 1 + 2

= n2 + 1

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 3. 1 + 3 + 7 + 13 + 21 + 31 +  . . . . n terms

Solution:

We are given the series: 1 + 3 + 7 + 13 + 21 + 31 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 1 + 3 + 7 + 13 + 21 + 31 +  . . . . + an–1 + an  . . . .(1)

S = 1 + 3 + 7 + 13 + 21 + 31 +  . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [1 + (3 + 7 + 13 + 21 + 31 +  . . . . + an–1 + an)] – [(1 + 3 + 7 + 13 + 21 + 31 +  . . . . + an–2 + an–1) + an]

=> 0 = 1 + [(3 – 1) + (7 – 3) + (13 – 7) + . . . . + (an – an–1)] – an

=> an = 1 + [2 + 4 + 6 + . . . . (n–1) terms]

As the series 2 + 4 + 6 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 4–2 = 2. So, we get,

=> an = 1 + (n–1) [2(2)+(n–2)2]/2

= 1+ (n–1) [4+2n–4]/2

= 1 + n(n–1)

= n2 – n+1

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 4. 3 + 7 + 14 + 24 + 37 +  . . . . n terms

Solution:

We are given the series: 3 + 7 + 14 + 24 + 37 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 3 + 7 + 14 + 24 + 37 +  . . . . + an–1 + a . . . .(1)

S = 3 + 7 + 14 + 24 + 37 +  . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [3 + (7 + 14 + 24 + 37 +  . . . . + an–1 + an)] – [(3 + 7 + 14 + 24 + 37 +  . . . . + an–2 + an–1) + an]

=> 0 = 3 + [(7 – 3) + (14 – 7) + (24 – 14) + . . . . + (an – an–1)] – an

=> an = 3 + [4 + 7 + 10 + . . . . (n–1) terms]

As the series 4 + 7 + 10 + . . . . (n–1) terms is an A.P., with first term(a) = 4 and common difference(d) = 7–4 = 3. So, we get,

=> an = 3 + (n–1) [2(4)+(n–2)3]/2

= 3+ (n–1) [8+3n–6]/2

= 3 + (n–1)(3n+2)/2

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 5. 1 + 3 + 6 + 10 + 15 +  . . . . n terms

Solution:

We are given the series: 1 + 3 + 6 + 10 + 15 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 1 + 3 + 6 + 10 + 15 +  . . . . + an–1 + an  . . . .(1)

S = 1 + 3 + 6 + 10 + 15 +  . . . . + an–2 + an–1 + a . . . .(2)

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

=> S – S = [1 + (3 + 6 + 10 + 15 +  . . . . + an–1 + an)] – [(1 + 3 + 6 + 10 + 15 +  . . . . + an–2 + an–1) + an]

=> 0 = 1 + [(3 – 1) + (6 – 3) + (10 – 6) + . . . . + (an – an–1)] – an

=> an = 1 + [2 + 3 + 4 + . . . . (n–1) terms]

As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,

=> an = 1 + (n–1) [2(2)+(n–2)1]/2

= 1+ (n–1) [4+n–2]/2

= 1 + (n–1)(n+2)/2

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 6. 1 + 4 + 13 + 40 + 121 +  . . . . n terms

Solution:

We are given the series: 1 + 4 + 13 + 40 + 121 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 1 + 4 + 13 + 40 + 121 +  . . . . + an–1 + an  . . . .(1)

S = 1 + 4 + 13 + 40 + 121 +  . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [1 + (4 + 13 + 40 + 121 +  . . . . + an–1 + an)] – [(1 + 4 + 13 + 40 + 121 +  . . . . + an–2 + an–1) + an]

=> 0 = 1 + [(4 – 1) + (13 – 4) + (40 – 13) + . . . . + (an – an–1)] – an

=> an = 1 + [3 + 9 + 27 + . . . . (n–1) terms]

As the series 3 + 9 + 27 + . . . . (n–1) terms is a G.P., with first term(a) = 3 and common ratio(r) = 9/3 = 3. So, we get,

=> an = 1 + 3(3n-1–1)/(3–1)

= 1+ 3(3n-1–1)/2

= 1 + 3n/2 – 3/2

= 3n/2 – 1/2

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 7. 4 + 6 + 9 + 13 + 18 +  . . . . n terms

Solution:

We are given the series: 4 + 6 + 9 + 13 + 18 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 4 + 6 + 9 + 13 + 18 +  . . . . + an–1 + an  . . . .(1)

S = 4 + 6 + 9 + 13 + 18 +  . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [4 + (6 + 9 + 13 + 18 +  . . . . + an–1 + an)] – [(4 + 6 + 9 + 13 + 18 +  . . . . + an–2 + an–1) + an]

=> 0 = 4 + [(6 – 4) + (9 – 6) + (13 – 9) + . . . . + (an – an–1)] – an

=> an = 4 + [2 + 3 + 4 + . . . . (n–1) terms]

As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,

=> an = 4 + (n–1) [2(2)+(n–2)1]/2

= 4 + (n–1)(n+2)/2

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 8. 2 + 4 + 7 + 11 + 16 +  . . . . n terms

Solution:

We are given the series: 2 + 4 + 7 + 11 + 16 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 2 + 4 + 7 + 11 + 16 +  . . . . + an–1 + an  . . . .(1)

S = 2 + 4 + 7 + 11 + 16 +  . . . . + an–2 + an–1 + an  . . . .(2)

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

=> S – S = [2 + (4 + 7 + 11 + 16 +  . . . . + an–1 + an)] – [(2 + 4 + 7 + 11 + 16 +  . . . . + an–2 + an–1) + an]

=> 0 = 2 + [(4 – 2) + (7 – 4) + (11 – 7) + . . . . + (an – an–1)] – an

=> an = 2 + [2 + 3 + 4 + . . . . (n–1) terms]

As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,

=> an = 2 + (n–1) [2(2)+(n–2)1]/2

= 2 + (n–1)(n+2)/2

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 9.

Solution:

The nth term of the given series would be,

an

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

### Question 10. to n terms.

Solution:

We are given the nth term of the given series,

an

Now we have to summate our nth term to find the sum(Sn) of this series.

Therefore, sum of the given series up to n terms is .

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