Find the sum of first 24 terms of an A-P- a 1- a 2- a 3- when a 1- a 5- a 10- a 15- a 20- a 24-225 is equal to 225

A_1+a_5+a_10+a_15+a_20+a_24=225 Sum of 1st 24 terms in AP is given by S_24=n/2[2a+(n 1)d] =24/2[2a_1+(24 1)d] =12[2a_1+23d] ∴ S_24=12(2a_1+23d) — (1) For ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Find the sum ...

Question

Find the sum of first 24 terms of an A.P. $a_{1}, a_{2}, a_{3} \dots$ ,when $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ is equal to 225

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Solution

$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
Sum of 1st 24 terms in AP is given by
$S_{24} = \frac{n}{2} [2 a + (n - 1) d] = \frac{24}{2} [2 a_{1} + (24 - 1) d] = 12 [2 a_{1} + 23 d]$
$∴ S_{24} = 12 (2 a_{1} + 23 d) - - - - (1) F o r A P, a_{n} = a + (n - 1) d$ .
$\begin{matrix} a_{1} = a_{1} & a_{15} = a_{1} + 14 d a_{5} = a_{1} + 4 d & a_{20} = a_{1} + 19 d a_{10} = a_{1} + 9 d & a_{24} = a_{1} + 23 d \end{matrix}$
$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$a_{1} + a_{1} + 4 d + a_{1} + 9 d + a_{1} + 14 d + a_{1} + 19 d + a_{1} + 23 d$ =225
$∴ 6 a + 69 d = 225$
3(2a+23d)=225
2a, +23d = 75 ---- (2)
$∴ S_{24} = 12 (75)$ [From (1) & (2)]
$S_{24} = 900$

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