Question

if the sum of first 9 terms of an a.p is 81 and that of 15 terms 225 find the sum of first 24 terms​

Answer 1

Given:-

• Sum of first 9 terms of an A.P = 81

• Sum of first 15 terms of an A.P = 225

To Find:-

• Sum of first 24 terms

Formula Used:-

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

Here,

• $\bf S_n$ = Sum of n terms

• a = First term of the A.P

• d = Common Difference

• n = No. of terms in an A.P

Solution:-

According to question,

$\bf :\implies\:S_9 = 81$

Using Formula,

$\bf :\implies\:S_n = \dfrac{n}{2}[2a+(n-1)d]$

Putting values,

$\sf :\implies\:81 = \dfrac{9}{2}[2a+(9-1)d]$

$\sf :\implies\:81\times2= 9[2a+8d]$

$\sf :\implies\:162= 9[2a+8d]$

$\sf :\implies\:2a+8d=\dfrac{162}{9}$

$\sf :\implies\:2a+8d=18$

$\sf :\implies\:a+4d=9$

$\sf :\implies\:a=9-4d$ ____{1}

And,

$\bf :\implies\:S_{15}= 225$

Using Formula,

$\bf :\implies\:S_n = \dfrac{n}{2}[2a+(n-1)d]$

Putting values,

$\sf :\implies\:225 = \dfrac{15}{2}[2a+(15-1)d]$

$\sf :\implies\:225\times2= 15[2a+14d]$

$\sf :\implies\:450= 15[2a+14d]$

$\sf :\implies\:2a+14d=\dfrac{450}{15}$

$\sf :\implies\:2a+14d=30$

$\sf :\implies\:a+7d=15$

Putting value of a,

$\sf :\implies\:9-4d+7d=15$

$\sf :\implies\:9+3d=15$

$\sf :\implies\:3d=15-9$

$\sf :\implies\:3d=6$

$\sf :\implies\:d=\dfrac{6}{3}$

$\sf :\implies\:d=2$

Putting value of d in {1},

$\sf :\implies\:a=9-4\times 2$

$\sf :\implies\:a=9-8$

$\sf :\implies\:a=1$

Now, Sum of First 24 terms:-

$\sf :\implies\:S_{24} = \dfrac{24}{2}[2\times 1+(24-1)2]$

$\sf :\implies\:S_{24} = 12[2+23\times 2]$

$\sf :\implies\:S_{24}= 12[2+46]$

$\sf :\implies\:S_{24} = 12\times 48$

$\sf :\implies\:S_{24} = 576$

Hence, The Sum of First 24 terms is 576.

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