iarc6io

2022-07-19

The first term and the last term of an AP are 17 and 350 respectively. If the common difference is 9,how many terms are there and what is their sum?

Izabelle Frost

Expert

Step 1
Given
First term of A.P $\left({a}_{1}\right)=17$
Last term of A.P $\left({a}_{n}\right)=350$
Common defference $\left(d\right)=9$
Step 2
finding the number of terms and their sum
${a}_{n}=350$
${a}_{1}+\left(n-1\right)d=350$
$17+\left(n-1\right)9=350$
$\left(n-1\right)9=350-17$
$\left(n-1\right)9=333$
$\left(n-1\right)=\frac{333}{9}$
$n-1=37$
$=37+1$
$=37+1$
there are 38 terns in A.P
WKT
Some to n terns ${S}_{n}=\frac{n}{2}\left[a+1\right]$
$=\frac{38}{2}\left[17+350\right]$
$=19\left[367\right]=6973$
The sum to 38 terms=6973

Glenn Hopkins

Expert

Given:
First term $\left(a\right)=17$
Last term $\left(l\right)={a}_{n}=350$
and common difference $\left(d\right)=9$
$\because {a}_{n}=350$
$a+\left(n-1\right)d=350$
$⇒17+\left(n-1\right)9=350$
$⇒9\left(n-1\right)=350-17=333$
$⇒n-1=\frac{333}{9}=37$
$\therefore n=38$
Now, ${S}_{n}=\frac{n}{2}\left(a+1\right)$
$=\frac{38}{2}\left(17+350\right)$
$=19×367=6973$
Hence, $n=38$ & sum of term ${S}_{n}=6973$

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