The first term of a G.P. is 7,the last term i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: First term $a = 7$,last term $l = a r^{n-1} = 448$,and sum of $n$ terms $S_n = 889$.The formula for the sum of $n$ terms of a $G.P.$ is $S_

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given: First term $a = 7$,last term $l = a r^{n-1} = 448$,and sum of $n$ terms $S_n = 889$.The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.We can rewrite this as $S_n = \frac{a r^n - a}{r - 1} = \frac{(a r^{n-1})r - a}{r - 1}$.Substituting the given values: $889 = \frac{448r - 7}{r - 1}$.Multiplying both sides by $(r - 1)$: $889(r - 1) = 448r - 7$.$889r - 889 = 448r - 7$.$889r - 448r = 889 - 7$.$441r = 882$.$r = \frac{882}{441} = 2$.Therefore,the common ratio is $2$.

The first term of a $G.P.$ is $7$,the last term is $448$,and the sum of all terms is $889$. Then the common ratio is:

Similar Questions

If $1 + \sin x + \sin^2 x + \dots \infty = 4 + 2\sqrt{3}$,where $0 < x < \pi$,then:

If the $19^{th}$ term of a non-zero $A.P.$ is zero,then the ratio of its ($49^{th}$ term) to ($29^{th}$ term) is:

The first and last terms of an $AP$ are $75$ and $375$ respectively. If its sum is $3600$,find the number of terms in the $AP$ and its common difference.

The difference between the fourth term and the first term of a Geometric Progression is $52.$ If the sum of its first three terms is $26,$ then the sum of the first six terms of the progression is

Let ${T_r}$ be the ${r^{th}}$ term of an $A.P.$ for $r = 1, 2, 3, ....$. If for some positive integers $m, n$ we have ${T_m} = \frac{1}{n}$ and ${T_n} = \frac{1}{m}$,then ${T_{mn}}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module