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. ⇒ [tex]\frac{a(r^n - 1)}{r-1}[/tex].
[NOTE :- [tex]\frac{a(1-r^n)}{1-r}[/tex] 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 :-
[tex]=> ar^3 = 384[/tex]
[tex]=> 6r^3 = 384[/tex]
Divide both the sides by 6.
[tex]=> \frac{6r^3}{6} = \frac{384}{6}[/tex]
[tex]=> r^3 = 64[/tex]
[tex]=> r = \sqrt[3]{64} = 4[/tex]
Sum of first 10 terms [tex]= \frac{6(4^{10}-1)}{4 - 1}[/tex]
[tex]= \frac{6(1048576 - 1)}{3}[/tex]
[tex]= 2 \times 1048575[/tex]
[tex]= 2097150[/tex]
