The sum of a few terms of a geometric progres | vedclass

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

Question

(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.The $n^{th}$ term of a geometric progression is given by $T_n = a r^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.The $n^{th}$ term of a geometric progression is given by $T_n = a r^{n-1}$.Given $T_n = 486$,we have $a(3)^{n-1} = 486$. Multiplying both sides by $3$,we get $a(3)^n = 1458$ .....$(i)$.The sum of $n$ terms is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given $S_n = 728$ and $r = 3$,we have $728 = \frac{a(3^n - 1)}{3 - 1}$.$728 = \frac{a(3^n) - a}{2}$.$1456 = a(3^n) - a$ .....$(ii)$.Substituting the value of $a(3^n)$ from equation $(i)$ into equation $(ii)$:$1456 = 1458 - a$.$a = 1458 - 1456 = 2$.Thus,the first term is $2$.

The sum of a few terms of a geometric progression is $728$. If the common ratio is $3$ and the last term is $486$,then the first term of the series will be:

Similar Questions

If the arithmetic mean of two numbers $a$ and $b$,$a > b > 0$,is five times their geometric mean,then $\frac{a + b}{a - b}$ is equal to

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a $G$.$P$. such that $a_{1} < 0$; $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$. If $\sum_{i=1}^{9} a_{i} = 4 \lambda$,then $\lambda$ is equal to:

If the sides of a right-angled triangle are in $A.P.$,then the sides are proportional to

The number of terms common to the two arithmetic progressions $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ is

In a $H.P.$,the $p^{th}$ term is $q$ and the $q^{th}$ term is $p$. Then the $pq^{th}$ term is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module