Arithmetic progression

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Question

If the first term of A.P is 2 and the sum of first five terms is equal to one fourth of the sum of the next five terms then the sum of first 30 terms is

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Solution

$T_{1} + T_{2} + T_{3} + T_{4} + T_{5} = \frac{1}{4} [T_{6} + T_{7} + T_{8} + T_{9} + T_{10}]$
$\begin{array}{l} \frac{5}{2} (T_{1} + T_{\underset{}{5}}) = \frac{1}{4} \cdot \frac{5}{2} (T_{6} + T_{10}) \\ \to 2 a + 4 d = \frac{1}{4} (2 a + 14 d) (\sin c e t_{n} = a + (n - 1) d) \\ \to d = - 3 a = - 3 (2) = - 6 \\ (\sin c e a = 2) \\ S_{30} = \frac{30}{2} [2 (2) + (30 - 1) (- 6)] = 15 x - 170 = - 2550 \\ (\sin c e s_{n} = \frac{n}{2} (2 a + (n - 1) d)) \end{array}$

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Arithmetic progression

Remember concepts with our Masterclasses.

If the first term of A.P is 2 and the sum of first five terms is equal to one fourth of the sum of the next five terms then the sum of first 30 terms is

T1+T2+T3+T4+T5=14T6+T7+T8+T9+T1052T1+T5 =14⋅52T6+T10→2a+4d=14(2a+14d) since tn=a+n-1d→d=−3a=−32=−6 since a=2S30=30222+(30−1)(−6)=15x−170=−2550 since sn=n22a+n-1d

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