Determine the sum of the first 35 terms of an | vedclass

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

Question

(B) Let $a$ be the first term and $d$ be the common difference of the $A.P.$We know that the $n^{th}$ term is given by $t_{n} = a + (n - 1)d$.Given $t

Vedclass · Accepted Answer

$2310$

Vedclass · Answer

(B) Let $a$ be the first term and $d$ be the common difference of the $A.P.$We know that the $n^{th}$ term is given by $t_{n} = a + (n - 1)d$.Given $t_{2} = 2$,we have $a + d = 2$ $...(1)$.Given $t_{7} = 22$,we have $a + 6d = 22$ $...(2)$.Subtracting equation $(1)$ from equation $(2)$:$(a + 6d) - (a + d) = 22 - 2$$5d = 20$,which gives $d = 4$.Substituting $d = 4$ in equation $(1)$:$a + 4 = 2$,so $a = -2$.The sum of the first $n$ terms is $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.For $n = 35$:$S_{35} = \frac{35}{2}[2(-2) + (35 - 1)4]$$S_{35} = \frac{35}{2}[-4 + 34 	imes 4]$$S_{35} = \frac{35}{2}[-4 + 136]$$S_{35} = \frac{35}{2} 	imes 132$$S_{35} = 35 	imes 66 = 2310$.

Determine the sum of the first $35$ terms of an $A.P.$ if $t_{2}=2$ and $t_{7}=22$.

Similar Questions

Find the $9^{th}$ term of the $H.P. 6, 4, 3, \ldots$

Terms $a, 1, b$ are in arithmetic progression and terms $1, a, b$ are in geometric progression. Find $a$ and $b$ (given $a \neq b$).

Let the sum of the first three terms of an $A.P.$ be $39$ and the sum of its last four terms be $178.$ If the first term of this $A.P.$ is $10,$ then the median of the $A.P.$ is

Write down the $5^{th}$ term of the series $\frac{1}{4} - \frac{1}{2} + 1 \ldots$

If $a, b, c, d$ are in $H.P.$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module