Question

find the first term of a g.p in which s6=4095and r =4​

Answer 1

Answer:-

Given:-

Sum of first 6 terms of a GP – S(6) = 4095

r = 4

We know that,

Sum of first n terms of a GP is given by:

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

Hence,

$\sf \implies \: S _{6} = \dfrac{a( {4}^{6 } - 1)}{4 - 1} \\ \\ \sf \implies \: 4095 = \dfrac{a(4096 - 1)}{3} \\ \\ \sf \implies \: \dfrac{4095}{3} \times a = 4095 \\ \\ \sf \implies \: a = 4095 \times \dfrac{3}{4095} \\ \\ \implies \red{ \sf \: a \: = \: 3}$

Therefore, the first term of the given GP is 3.

Some Important Formulae:

• nth term of a GP = $\sf a \times r^{n - 1}$

• Sum of an infinite GP = $\sf \dfrac{a}{1 - r}$

• Three numbers in GP are taken as a/r , a , ar.

• Four numbers in GP are a/r³ , a/r , ar , ar³.

• Five numbers in GP are a/r² , a/r , a , ar , ar².

Answer 2

Answer:

✡ Given ✡

⬛ A G .P. in which $\sf{S_6}$ = 4095 r= 4.

✡ To Find ✡

⬛ The first term ?

✡ Solution ✡

⬛ The sum formula of a G.P is $\sf{S_n}$ = $\dfrac{a({r^{n}}-1)}{r-1}$

Here,$\sf{S_n}$ = 4095, n = 6 and r = 2

✏ Substitute all ,

=> 4095 = $\dfrac{a({4^{6}}-1)}{4-1}$

=> 4095 = $\dfrac{a(4⁶-1)}{4-1}$

=> 4095 = $\dfrac{a(4096-1)}{3}$

=> a = $\dfrac{4095×3}{4095}$

=> a = 3

Therefore, the first term is 3

Step-by-step explanation:

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