Question

Find the sum of series 7 + 10 1/2+ 14 + ... .... + 84​

Answer 1

$\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}$

$\boxed{\implies 1046.5}$

$\boxed {\underline {\mathbb {GIVEN:-}}}$

$ap$ in series is $7+ 10\frac{1}{2}+14+..+84$

$\boxed {\underline {\mathbb {TO\:FIND:-}}}$

sum of series\ap

$\boxed {\underline {\mathbb {FORMULA\:USED:-}}}$

$a_n = a+(n-1)d$  

$S_n=\frac{n}{2}(a+l)$

$\boxed {\underline {\mathbb {SOLUTION:-}}}$

Here  

$ap$ in series is $7+ 10\frac{1}{2}+14+..+84$

or

$7,10\frac{1}{2},14..84$

As we have given $a$(first term)$=7$

So let’s find d(common difference)  

$14=7+2d\\14-7=2d\\7=2d$

$d=\frac{7}{2}$

now let’s get the n term of ap as last term is 84

$a_n = a+(n-1)d$  

$84=7+(n-1)\frac{7}{2}$

$84=7+\frac{7n}{2} - \frac{7}{2}$

$84=\frac{14+7n-7}{2}$

$84=\frac{7+7n}{2}$

$84 \times 2 = 7 +7n\\168=7 + 7n\\7n=168-7\\7n=161$

$n=\frac{161}{7}$

$n=23$

hence the ap have total of $23$ $terms$

Now let’s find $S_n$ (sum of ap)

$S_n=\frac{n}{2}(a+l)$

$S_n=\frac{23}{2}(7+84)$

$S_n=\frac{23}{2}(91)$

$S_n=\frac{23*91}{2}$

$S_n=1046.5$

Hence sum of the whole ap $\boxed{\implies 1046.5}$

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here

a=first term

d=common difference

$a_n= term\ which \ we \ want\ to\ find$

$S_n = sum\ of\ ap$

l= last term for of series

n= number of that term