Question

The first term of a geometric sequence is 6 and the fourth term is 384. What is the sum of the first 10 terms of the corresponding series?​

Answer 1

Answer:

2097150

Step-by-step explanation:

GIVEN :-

• First term of G.P. = 6
• Forth term of G.P. = 384

TO FIND :-

• Sum of first 10 terms of the G.P.

CONCEPT TO BE USED IN THIS QUESTION :-

Geometric Progression :-

• It's a sequence in which the successive terms have same ratio.

• General form of a G.P. ⇒ a , ar , ar² , ar³ , ....... [where a = first term ; r = common ratio between successive terms]

• Sum of 'n' terms of a G.P. ⇒ $\frac{a(r^n - 1)}{r-1}$.

[NOTE :- $\frac{a(1-r^n)}{1-r}$ can also be the formula for "Sum of n terms of G.P." because if you put 'r' there (assuming r > 0) you'll get negative value in both the numerator &  denominator from which the negative sign will get cancelled from the numerator & denominator. YOU'LL BE GETTING THE SAME VALUE FROM BOTH THE FORMULAES.]

SOLUTION :-

Let the first term of the G.P. given in the question be 'a' and the common ratio between successive terms be 'r'.

⇒ a = 6

It's given that forth term is 384. So from "General form of G.P." , it can be stated that :-

$=> ar^3 = 384$

$=> 6r^3 = 384$

Divide both the sides by 6.

$=> \frac{6r^3}{6} = \frac{384}{6}$

$=> r^3 = 64$

$=> r = \sqrt[3]{64} = 4$

Sum of first 10 terms $= \frac{6(4^{10}-1)}{4 - 1}$

                                     $= \frac{6(1048576 - 1)}{3}$

                                     $= 2 \times 1048575$

                                     $= 2097150$