Question

-If the sum of first 8 terms of an AP is 64 and the sum of first 12 terms is144, find the sum of n terms of an AP. ​

Answer 1

GIVEN ;

        s8 = 4 (a +7 d )

         s12 = 6 (a +11d )

64 = 4a + 28d

144= 6a + 66d

a/q

   a = a

(64 - 28d)/4 = (144 -66d)/6

=> 384 - 168 d = 576 - 266d

=> 98 d = 192

d = 192 /98

therefore

              sn = n/2 ( a + (n-1)192/98)

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Answer 2

Solution :

$\bigstar$ We know that formula of the sum of an A.P;

$\boxed{\bf{S_n=\frac{n}{2}\bigg[2a+(n-1)d\bigg]}}}$

• n is the term of an A.P.
• a is the first term.
• d is the common difference.

A/q

$\longrightarrow\sf{S_8=64}\\\\\longrightarrow\sf{\cancel{\dfrac{8}{2}} \bigg[2a+(8-1)d\bigg]=64}\\\\\longrightarrow\sf{4[2a+7d]=64}\\\\\longrightarrow\sf{2a+7d=\cancel{64/4}}\\\\\longrightarrow\sf{2a+7d=16.....................(1)}$

&

$\longrightarrow\sf{S_{12}=144}\\\\\longrightarrow\sf{\cancel{\dfrac{12}{2}} \bigg[2a+(12-1)d\bigg]=144}\\\\\longrightarrow\sf{6[2a+11d]=144}\\\\\longrightarrow\sf{2a+11d=\cancel{144/6}}\\\\\longrightarrow\sf{2a+11d=24.....................(2)}$

∴Subtracting equation (1) & equation (2),we get;

$\longrightarrow\sf{\cancel{2a-2a}+7d-11d=16-24}\\\\\longrightarrow\sf{-4d=-8}\\\\\longrightarrow\sf{d=\cancel{-8/-4}}\\\\\longrightarrow\bf{d=2}$

Putting the value of d in equation (1),we get;

$\longrightarrow\sf{2a+7(2)=16}\\\\\longrightarrow\sf{2a+14=16}\\\\\longrightarrow\sf{2a=16-14}\\\\\longrightarrow\sf{2a=2}\\\\\longrightarrow\sf{a=\cancel{2/2}}\\\\\longrightarrow\bf{a=1}$

Now;

$\longrightarrow\sf{S_n=\dfrac{n}{2} \bigg[2a+(n-1)d\bigg]}\\\\\\\longrightarrow\sf{S_n=\dfrac{n}{2} \bigg[2(1)+(n-1)(2)\bigg]}\\\\\\\longrightarrow\sf{S_n=\dfrac{n}{2} \bigg[2+2n-2\bigg]}\\\\\\\longrightarrow\sf{S_n=\dfrac{n}{2} \bigg[2n\cancel{+2-2}\bigg]}\\\\\\\longrightarrow\sf{S_n=\dfrac{n}{\cancel{2}} \times \cancel{2}n}\\\\\longrightarrow\sf{S_n=n\times n}\\\\\longrightarrow\bf{S_n=n^{2} }$

Thus;

The sum of nth term of an A.P. will be n² .