Question

consider an arithmetic sequence whose 6th term is 40and 9th term is 58 a)find the sum of first 25 terms of the sequence​

Answer 1

Given:

• In an A.P. 6th term is 40 and 9th term is 58 .

To find:

• The sum of first 25 terms of the sequence.

Solution:

$\boxed{\it{tn=a+(n-1)d}} \\ \\ \sf{According \ to \ the \ fist \ condition} \\ \\ \sf{a+5d=40...(1)} \\ \\ \sf{According \ to \ the \ second \ condition} \\ \\ \sf{a+8d=58...(2)} \\ \\ \sf{Subtract \ eq(1) \ from \ eq(2), \ we \ get} \\ \\ \sf{3d=18} \\ \\ \boxed{\sf{\therefore{d=6}}} \\ \\ \sf{Substitute \ d=6 \ in \ eq(1), \ we \ get} \\ \\ \sf{a+5(6)=40} \\ \\ \boxed{\sf{\therefore{a=10}}} \\ \\ \boxed{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}} \\ \\ \sf{\leadsto{S_{25}=\dfrac{25}{2}[2(10)+(25-1)6]}} \\ \\ \sf{\leadsto{S_{25}=12.5[20+144]}} \\ \\ \sf{\leadsto{S_{25}=2050}} \\ \\ \sf{\therefore{Sum \ of \ first \ 25 \ terms \ is \ 2050.}}$