   Question Answers

# In an A.P $a = 7,{a_{13}} = 35$ ,find$d$ and${S_{13}}$  Hint: Here, we are going to use (i) the formula for the nth term of an A.P
(ii) the formula for them sum till the nth term

Given, $a = 7,{a_{13}} = 35$
We know ${a_n} = a + (n - 1)d$
$\Rightarrow {a_{13}} = 7 + (13 - 1)d$
$\Rightarrow 35 = 7 + 12d$
$\Rightarrow 28 = 12d$
$\Rightarrow d = \dfrac{{28}}{{12}}$
$\Rightarrow d = \dfrac{7}{3}$
Now, the sum up to nth terms is given by ${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {2 \times 7 + (13 - 1)\dfrac{7}{3}} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {14 + 12 \times \dfrac{7}{3}} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {14 + 4 \times 7} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {14 + 28} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {42} \right)$
$\Rightarrow {S_{13}} = 13 \times 24$
$\Rightarrow {S_{13}} = 312$
Therefore, $d = \dfrac{7}{3}$and${S_{13}} = 312$are the required solution

Note:(i) One has to find the common difference first, in order to calculate the sum to nth term.
(ii) One should not confuse between if he/she has to find the nth from the beginning or from the end. If nothing has been mentioned in the question then by default you have to find it from the beginning.

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