Find the sum of $24$ terms of arithmetic progression if $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24}$ is $225$

800

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900

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700

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650

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Solution

The correct option is B $900$
Let first term $= a,$ common difference $= d$ and $n^{t h}$ term $= a n = a + (n - 1) d$ .
So, $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225 a + a + 4 d + a + 9 d + a + 14 d + a + 19 d + a + 23 d = 225 6 a + 69 d = 225 3 (2 a + 23 d) = 225 2 a + 23 d = 75$

and, $S_{24} = \frac{24}{2} [2 a + 23 d] = 12 (2 a + 23 d) = 12 \times 75 = 900$

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Similar questions

Q. If

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Find the sum of 24 terms of arithmetic progression if a1+a5+a10+a15+a20+a24 is 225

The correct option is B 900Let first term =a, common difference =d and nth term =an=a+(n−1)d.So, a1+a5+a10+a15+a20+a24=225a+a+4d+a+9d+a+14d+a+19d+a+23d=2256a+69d=2253(2a+23d)=2252a+23d=75and, S24=242[2a+23d]=12(2a+23d)=12×75=900

Find the sum of $24$ terms of arithmetic progression if $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24}$ is $225$