Let S_n denote the sum of the first n terms of an arithmetic progression. If S_10 = 390 and the ratio of the tenth and the fifth terms is 15:7, then S_15 - S_5 is equal to:

(c): We have, S_10=390 \ => 5(2 a+9 d)=390 => 2 a+9 d=78 (i) Also, fraca_10a_5=(15 / 7) => (a+9 d / a+4 d)=(15 / 7) \ => 7 a+63 d=15 a+60 d \ => 8 a=3 d \ => a=(3 / 8) d (ii) From (i) (ii), we get d=8 and a=3 Now, S_15-S_5=(15 / 2)[6+112]-(5 / 2)[6+32]=790

Let S_{n} denote the sum of the first n terms of an ari... | Sequence & Series (Mathematics)

(c): We have, $S_{10}=390$ $\\ \Rightarrow 5(2 a+9 d)=390 \Rightarrow 2 a+9 d=78\\$ (i) Also, $\frac{a_{10}}{a_5}=\frac{15}{7} \Rightarrow \frac{a+9 d}{a+4 d}=\frac{15}{7}$ $\\ \Rightarrow 7 a+63 d=15 a+60 d \\ \Rightarrow 8 a=3 d$ $\\ \Rightarrow a=\frac{3}{8}$ d (ii) $\\$ From (i) & (ii), we get $d=8$ and $a=3\\$ Now, $S_{15}-S_5=\frac{15}{2}[6+112]-\frac{5}{2}[6+32]=790$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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