Question

The sum of first 20 terms common between the series $3 + 7 + 11 + 15.....$ and $1 + 6 + 11 + 16....$ is
A. $4020$
B. $4220$
C. $4200$
D. $4000$

Answer 1

We find the common difference of both the Arithmetic progression which is written as series. Then we write a few terms of both series to find two common terms which will form our new series of common terms. Using the two terms in the new series we find the common difference of the new series and find its sum using the formula for the sum of n terms.

• In an AP, first term is a, common difference is d and the number of term is n, then the $n^{th}$ term is given by ${a_n} = a + (n - 1)d$
• The sum of n terms of an AP is given by ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$

Complete step-by-step answer:
Now we have the first series as $3 + 7 + 11 + 15.....$
We can see the terms of the series form an AP $3,7,11,15.....$ where

$$a = 3 \\\ d = 4 \\\$$

We can write the series in an extended form as $3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35....$
Now we have the second series as $1 + 6 + 11 + 16....$
We can see the terms of the series form an AP $1,6,11,16....$ where

$$a = 1 \\\ d = 5 \\\$$

We can write the series in an extended form as $1 + 6 + 11 + 16 + 21 + 26 + 31 + 36....$
Looking at both series we can see two common terms in the starting as $11,31$.
We take these two terms and form a new series which will have common difference $31 - 11 = 20$
So, the new series is $11 + 31 + 51 + 71......$ where

$$a = 11 \\\ d = 20 \\\$$

We have to find sum of first 20 terms, so we put n as 20 in the formula for sum of n terms, i.e. ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$.
$\Rightarrow {S_{20}} = \dfrac{{20}}{2}(2 \times 11 + (20 - 1)20)$
Cancel the common factors from numerator and denominator and solve the value in the bracket

$$\Rightarrow {S_{20}} = 10(22 + (19) \times 20) \\\ \Rightarrow {S_{20}} = 10(22 + 380) \\\ \Rightarrow {S_{20}} = 10 \times 402 \\\ \Rightarrow {S_{20}} = 4020 \\\$$

So, the sum of the first 20 terms of the series formed by common terms of two given series is 4020.

So, the correct answer is “Option A”.

Note: Students are likely to make mistakes if they don’t understand the language of the question and they calculate the sum of the first 20 terms of both series and proceed with a solution which is wrong.