The sum of first $24 t h$ terms of the A.P $a_{1}, a_{2}, a_{3} . . . . . . . .$ if it is known that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ , is

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Solution

Given $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225 \Rightarrow a + a + 4 d + a + 9 d + a + 14 d + a + 19 d + a + 19 d + a + 23 d = 225 \Rightarrow 6 a + 69 d = 225$
$\Rightarrow 2 a + 23 d = 75$ (where $a$ is the first term & $d$ is the common difference of $A P$ )
$S_{24} = \frac{n}{2} [2 a + (n - 1) d]$
$= \frac{24}{2} [2 a + (23) d] = 12 \times (2 a + 23 d) = 12 \times 75 = 900$

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Similar questions

a_{1}, a_{2}, a_{3}, . . . . . . . . .

A . P .

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

24

A . P

a_{1}, a_{2}, a_{3} \dots

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1}, a_{2}, a_{3}

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} =

a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24} =

If $a_{1}, a_{2}, a_{3} . . . . .$ are in $A . P .$ such that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ .

a_{1} + a_{2} + a_{3} . . . . . . . . . + a_{23} + a_{24}

a_{1}, a_{2}, a_{3}, . . . .

a_{1}, a_{5}, a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . + a_{24}

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