Question

the first term of an ap is 10 last term is -6 and number of term is 5 find the common difference and sum of the series described this question solution​

Answer 1

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

$\sf{The \ common \ difference \ and \ the \ sum \ of}$

$\sf{the \ series \ are \ -4 \ and \ 10 \ respectively.}$

$\sf\orange{Given:}$

$\sf{In \ an \ AP,}$

$\sf{\implies{First \ term \ (a)=10}}$

$\sf{\implies{Number \ of \ terms=5}}$

$\sf{\implies{Last \ term(t5)=-6}}$

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

$\sf{Common \ difference \ and \ sum \ of \ the \ series.}$

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

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

$\sf{\therefore{-6=10+(5-1)d}}$

$\sf{\therefore{4d=-6-10}}$

$\sf{\therefore{4d=-16}}$

$\sf{\therefore{d=\frac{-16}{4}}}$

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

$\boxed{\sf{S_{n}=\frac{n}{2}[t_{1}+t_{n}]}}$

$\sf{\therefore{S_{5}=\frac{5}{2}[10+(-6)]}}$

$\sf{\therefore{S_{5}=\frac{5}{2}\times4}}$

$\sf{\therefore{S_{5}=5\times2}}$

$\sf{\therefore{S_{5}=10}}$

$\sf\purple{\tt{\therefore{The \ common \ difference \ and \ the \ sum \ of}}}$

$\sf\purple{\tt{the \ series \ are \ -4 \ and \ 10 \ respectively.}}$

Answer 2

Answer:

d = -4 and Sum of 5 terms is 20

Refer the attached picture

$\green{THANK YOU}$