If the first,second and last terms of an A.P. | vedclass

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

Question

(C) Given the first term $A = a$ and the second term $A + d = b$.Common difference $d = b - a$.The last term $l = 2a$. The formula for the last term i

Vedclass · Accepted Answer

$\frac{3ab}{2(b - a)}$

Vedclass · Answer

(C) Given the first term $A = a$ and the second term $A + d = b$.Common difference $d = b - a$.The last term $l = 2a$. The formula for the last term is $l = A + (n - 1)d$.Substituting the values: $2a = a + (n - 1)(b - a)$.$a = (n - 1)(b - a) \implies n - 1 = \frac{a}{b - a} \implies n = \frac{a}{b - a} + 1 = \frac{a + b - a}{b - a} = \frac{b}{b - a}$.The sum of an $A.P.$ is $S_n = \frac{n}{2}(A + l)$.$S_n = \frac{b}{2(b - a)}(a + 2a) = \frac{b}{2(b - a)}(3a) = \frac{3ab}{2(b - a)}$.

If the first,second and last terms of an $A.P.$ are $a, b$ and $2a$ respectively,then its sum is:

Similar Questions

If the $p^{th}$ term of an arithmetic progression is $q$ and its $q^{th}$ term is $p$,then what is its $(p + q)^{th}$ term?

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

Let ${a_1}, {a_2}, \dots, {a_{49}}$ be in $A.P.$ such that $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $\sum_{r = 1}^{17} a_r^2 = 140m$,then $m = \dots$

If the $n^{th}$ term of an arithmetic progression is $\frac{(2n + 1)}{3}$,what is the sum of its first $19$ terms?

If $a, b, c, d, e$ are in $A.P.$,then the value of $a + b + 4c - 4d + e$ in terms of $a$,if possible,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module