Question

If $a_1,a_2,......,a_{24}$ are in AP and $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$ then the sum of 24 terms of this AP is.

• A. $900$
• B. $450$
• C. $225$
• D. None of these

Answer 1

The correct option is A $900$

Given; $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$

$(a_1+a_{24})+(a_5+a_{20})+(a_{10}+a_{15})=225$ ...(1)

Let first term is $a$ and common difference is $d$ then

$a_1=a,a_{24}=a+23d,\Rightarrow a_1+a_{24}=2a+23d$ ...(2)

$a_5=a+4d,a_{20}=a+19d,\Rightarrow a_5+a_{20}=2a+23d$ ...(3)

$a_{10}=a+9d,a_{15}=a+14d,\Rightarrow a_{10}+a_{15}=2a+23d$ ...(4)

Putting (2),(3),(4) in (1), we get

$3(2a+23d)=225\Rightarrow 2a+23d=75$ ...(5)

Now $a_1+a_2+a_3+a_4+...+a_{24}=524$

$\Rightarrow 524=\frac{24}{2}[2a+(24-1)d]=12(2a+23d)$ ...(6)

From (5) and (6)

$S_{24}=12\times 75=900$