Question

the sum of the frst three terms of an AP is 42 and the product of the frst and third term is 52 find the first term and common difference also write first terms

Answer 1

ANSWER:

Given:

• Sum of first 3 terms of an AP = 42
• Product of first and third term = 52

To Find:

• First term
• Common difference
• Series.

Solution:

$\text{Let the terms be (a - d), a and (a + d).}\\\\\text{We are given that,}\\\\:\longrightarrow (a-d)+(a)+(a+d)=42\\\\:\implies a-d\!\!\!/+a+a+d\!\!\!/=42\\\\:\implies3a=42\\\\:\implies a=\dfrac{42\!\!\!\!/^{\:14}}{3\!\!\!/}\\\\\bf{:\implies a=14}.\\\\\text{Also,}\\\\:\longrightarrow(a-d)(a+d)=52\\\\\text{We know that,}\\\\:\hookrightarrow(x-y)(x+y)=x^2-y^2\\\\\text{So,}\\\\:\implies a^2-d^2=52$

$:\implies(14)^2-d^2=52\\\\:\implies196-d^2=52\\\\:\implies d^2=196-52\\\\:\implies d^2=144\\\\\bf{:\implies d=\pm12}\\\\\text{So,}\\\\\text{The series is,}\\\\\text{For d=+12}\\\\:\implies(a-d)=14-12=2\\\\:\implies a=14\\\\:\implies(a+d)=14+12=26\\\\\text{And,}\\\\\text{For d=+12}\\\\:\implies(a-d)=14+12=26\\\\:\implies a=14\\\\:\implies(a+d)=14-12=2$

$\bf{Hence,}\\\\\bf{1)\:First\:term(a)=14}\\\\\bf{2)\:Common\:difference(d)=\pm12}\\\\\bf{3)\:Series=26,14,2\:or\:2,14,26}$

Formula Used:

• $(x-y)(x+y)=x^2-y^2$