Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, $$\dfrac{1}{2}$$.

Asked by Pragya Singh | 1 year ago |  101

#### 1 Answer

##### Solution :-

The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x $$\dfrac{1}{2}$$

= 64[4 + 2 + 1 + $$\dfrac{1}{2}$$+ $$\dfrac{1}{2^2}$$]

Now, it’s seen that

4, 2, 1, $$\dfrac{1}{2}$$, $$\dfrac{1}{2^2}$$ is a G.P.

With first term, a = 4

Common ratio, r =$$\dfrac{1}{2}$$

We know,

$$s_n=\dfrac{a(1-r^n)}{1-r}$$

$$8( \dfrac{32-1}{32})=\dfrac{31}{4}$$

Answered by Abhisek | 1 year ago

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