Question

find the sum of the first 25 term of each of the arithmetic sequences below
11,22,33,...​

Answer 1

Answer:

a=11

d=11

n=25

s=25/2*(22+(25-1)11)

s=25/2*286

s=3075

Answer 2

Answer:

$\sf{The \ sum \ of \ first \ 25 \ terms \ is \ 3575.}$

Given:

$\textsf{The given arithmetic sequence is 11, 22, 33,...}$

To find:

$\sf{The \ sum \ of \ first \ 25 \ terms.}$

Solution:

$\sf{The \ given \ arithmetic \ sequence \ is \ 11, \ 22, \ 33,...} \\ \sf{Here, \ a=11 \ and \ d=22-11=11} \\ \\ \\ \boxed{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}} \\ \\ \sf{S_{25}=\dfrac{25}{2}[2(11)+(25-1)\times11]} \\ \\ \sf{\therefore{S_{25}=\dfrac{25}{2}[2(11)+24\times11]}} \\ \\ \sf{\therefore{S_{25}=25[11+12\times11]}} \\ \\ \sf{\therefore{S_{25}=25\times143}} \\ \\ \sf{\therefore{S_{25}=3575}} \\ \\ \purple{\tt{\therefore{The \ sum \ of \ first \ 25 \ terms \ is \ 3575.}}}$

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Extra information:

$\sf{\leadsto{t_{n}=a+(n-1)d}} \\ \\ \sf{\leadsto{S_{n}=\dfrac{n}{2}[t_{1}+t_{n}]}} \\ \\ \sf{\leadsto{2\times \ t_{2}=t_{1}+t_{3}}}$