Question

The sum of 1st and 17th term of an arithmetic sequence is 40 the sum of its 1st and 18th term 43

Answer 1

Correct Question :

• The sum of the 1st and 17th terms of an arithmetic sequence is 40.the sum of its 1st and 18th terms is 43.what is it's common difference.

AnSwer :

• The nth term of an AP is an = a1 + (n-1)d.

Similarly,

• The 16th term is a1 + 16d

Now,

• Sum of the 1st and 17th term is 2a1 + 16d = 40

For the 18th term, the sum required is 2a1 + 17d = 43

★ Difference (d) = 43 - 40= 3

Hence , the common difference of the given terms is 3.

Answer 2

$\sf\blue{Correct \ Question}$

$\sf{The \ sum \ of \ 1^{st} \ and \ 17^{th} \ term \ of}$

$\sf{an \ arithmetic \ sequence \ is \ 40 \ the \ sum}$

$\sf{of \ it's \ 1^{st} \ and \ 18^{th} \ term \ is \ 43.}$

$\sf{Find \ the \ first \ term \ and \ common \ difference.}$

___________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ first \ term \ is \ -4 \ and \ common}$

$\sf{difference \ is \ 3.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ sum \ of \ 1^{st} \ and \ 17^{th}}}$

$\sf{term \ is \ 40.}$

$\sf{\implies{The \ sum \ of \ 1^{st} \ and \ 18^{th}}}$

$\sf{term \ is \ 43.}$

$\sf\pink{To \ find:}$

$\sf{First \ term \ and \ common \ difference.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{a+a+16d=40}$

$\sf{\therefore{2a=40-16d...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{2a+17d=43...(2)}$

$\sf{Substitute \ 2a=40-16d \ in \ eq(2), \ we \ get}$

$\sf{\therefore{(40-16d)+17d=43}}$

$\sf{\therefore{40+d=43}}$

$\sf{\therefore{d=43-40}}$

$\sf{\therefore{d=3}}$

$\sf{Substitute \ d=3 \ in \ eq(1)}$

$\sf{2a=40-16(3)}$

$\sf{\therefore{a=40-48}}$

$\sf{\therefore{2a=-8}}$

$\sf{\therefore{a=\frac{-8}{2}}}$

$\sf{\therefore{a=-4}}$

$\sf\purple{\tt{\therefore{The \ first \ term \ is \ -4 \ and \ common}}}$

$\sf\purple{\tt{difference \ is \ 3.}}$