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

Minimum value of $2 \sin^2 \theta + 8 \csc^2 \theta$ is (where $\theta \in R$):-

The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms is:

If the harmonic mean between $a$ and $b$ is $H$,then the value of $\frac{1}{H - a} + \frac{1}{H - b}$ is

For any odd integer $n \ge 1$,${n^3} - {(n - 1)^3} + ........... + {( - 1)^{n - 1}}{1^3} = $

If $a_1, a_2, a_3, \dots, a_n$ are in $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module