Determine the 25^{th} term of an A.P. whose 9 | vedclass

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

Question

(C) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.Giv

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.Given that the $9^{th}$ term $a_9 = -6$ and the common difference $d = \frac{5}{4}$.Substituting these values into the formula:$a_9 = a + (9 - 1)d$$-6 = a + 8 	imes \left(\frac{5}{4}ight)$$-6 = a + 2 	imes 5$$-6 = a + 10$$a = -6 - 10 = -16$.Now,we need to find the $25^{th}$ term $(a_{25})$:$a_{25} = a + (25 - 1)d$$a_{25} = -16 + 24 	imes \left(\frac{5}{4}ight)$$a_{25} = -16 + 6 	imes 5$$a_{25} = -16 + 30 = 14$.Thus,the $25^{th}$ term is $14$.

Determine the $25^{th}$ term of an $A.P.$ whose $9^{th}$ term is $-6$ and the common difference is $\frac{5}{4}$.

Similar Questions

If $a_1 = a_2 = 2$ and $a_n = a_{n-1} - 1$ for $n > 2$,then $a_5$ is:

The sum of the series $1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$ is

Find the sum of the first $7$ terms of a $GP,$ whose first term is $1024$ and the common ratio is $\frac{1}{2}$.

Insert $6$ arithmetic means between $20$ and $76$.

The number $111...1$ ($91$ times) is a:

Vedclass Test Series

Exam Paper Generator

Online Exam Module