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Visual SolutionPYQ 2023 · Apr 6 Shift 1Standard

Question

The sum of the first 20 terms of the series

5 + 11 + 19 + 29 + 41 + \ldots

is

(A)

3520

(B)

3450

(C)

3250

(D)

3420

Solution Path

Differences form AP with

d=2

, so

T_n = n^2 + 3n + 1

. Then

S_{20} = 2870 + 630 + 20 = 3520

.

01Question Setup

1/4

Find the sum of first 20 terms of

5 + 11 + 19 + 29 + 41 + \ldots

S_{20} = \;?

02Method of Differences

2/4

First differences

6, 8, 10, 12, \ldots

form an AP with

d = 2

. General term:

T_n = n^2 + 3n + 1

.

T_n = n^2 + 3n + 1

03Sum to 20 TermsKEY INSIGHT

3/4

S_{20} = \sum(n^2 + 3n + 1) = \dfrac{20 \cdot 21 \cdot 41}{6} + 3 \cdot \dfrac{20 \cdot 21}{2} + 20 = 2870 + 630 + 20 = 3520

.

S_{20} = 3520

04Final Answer

4/4

S_{20} = 3520

.

\boxed{3520}

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