If the sum of three consecutive terms of an A | vedclass

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

Question

(A) Let the three consecutive terms of an $A.P.$ be $(a - d)$,$a$,and $(a + d)$.According to the given condition,the sum of these terms is $51$:$(a -

Vedclass · Accepted Answer

$21, 17, 13$

Vedclass · Answer

(A) Let the three consecutive terms of an $A.P.$ be $(a - d)$,$a$,and $(a + d)$.According to the given condition,the sum of these terms is $51$:$(a - d) + a + (a + d) = 51$$3a = 51$$a = 17$The product of the first term $(a - d)$ and the last term $(a + d)$ is $273$:$(a - d)(a + d) = 273$$a^2 - d^2 = 273$Substituting $a = 17$:$17^2 - d^2 = 273$$289 - d^2 = 273$$d^2 = 289 - 273 = 16$$d = \pm 4$If $d = 4$,the terms are $(17 - 4), 17, (17 + 4)$,which are $13, 17, 21$.If $d = -4$,the terms are $(17 - (-4)), 17, (17 + (-4))$,which are $21, 17, 13$.Both sets represent the same sequence. Thus,the numbers are $21, 17, 13$.

If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of the last and first term is $273$,then the numbers are

Similar Questions

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

Sum the series $51 + 50 + 49 + \ldots + 21$.

Let $f: R \rightarrow R$ be such that for all $x \in R$,$(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P.$. Then,the minimum value of $f(x)$ is:

If the numbers $a, b, c, d, e$ form an $A.P.$,then the value of $a - 4b + 6c - 4d + e$ is

If the sum of the first $n$ terms of two arithmetic progressions $3, 7, 11, 15, \ldots$ and $30, 33, 36, 39, \ldots$ are equal,then find the value of $n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module