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Question

Answers

A. \[4020\]

B. \[4220\]

C. \[4200\]

D. \[4000\]

Answer
Verified

* 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 \[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]

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.