Let S_n denote the sum of first n terms of an arithmeticn progression. If S_20 = 790 and S_10 = 145, thenn S_15 - S_5 is

(d) : Given, S_20=790nn => (20 / 2)(2 a+19 d)=790 => 2 a+19 d=79 \ (i)nnGiven, S_10=145nn(10 / 2)(2 a+9 d)=145 => 2 a+9 d=29 \ (i i)nnFrom (i) and (ii), we get 10 d=50 => d=5nFrom (i), we get a=(79-95 / 2)=-8nSo, S_15-S_5=(15 / 2)[-16+70]-(5 / 2)[-16+20]=395

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

(d) : Given,

S_{20}=790

\n\Rightarrow \frac{20}{2}(2 a+19 d)=790 \Rightarrow 2 a+19 d=79 \#(i)\n

\nGiven,

S_{10}=145

\n\frac{10}{2}(2 a+9 d)=145 \Rightarrow 2 a+9 d=29 \#(i i)\n

\nFrom (i) and (ii), we get

10 d=50 \Rightarrow d=5

\nFrom (i), we get

a=\frac{79-95}{2}=-8

\nSo,

S_{15}-S_5=\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20]=395

Explanation

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