Question

The sum of the first three terms of an AP is 42. If
the product of the first and the third term exceeds
the second term by 157, then the 10th term of the
AP can be
(1) 59
(2) 54
(3) 49
(4) 44

Answer 1

Answer:

2) 54

Step-by-step explanation:

First term of an AP= a

Second term of an AP= a+d

Third term of an AP= a+2d

and so on...

a-d+a+a+d= 42

3a=42

a=14

1st term is 14

(a-d)(a+d)= a+157

a^2-d^2= a+157

196-d^2= 171

196-171= d^2

25= d^2

d=5

10th term= a+8d

               = 14+40

                = 54

Answer 2

$\sf\large\underline{Let:}$

$\rm{\implies The\: terms\:of\:AP\:be\: a-d,\:a,\:a+d}$

$\sf\large\underline{To\:Find:}$

$\rm{\implies The\: 10th\:terms\:of\:AP=?}$

$\sf\large\underline{Solution:}$

$\sf\small\underline{Given\:in\:1st\:case:}$

The sum of 1st 3 terms=42

$\tt{\implies a-d+a+a+d=42}$

$\tt{\implies 3a=42}$

$\tt{\implies a=14}$

$\sf\small\underline{Given\:in\:2nd\:case:}$

the product of the first and the third term exceeds the second term by 157]

$\tt{\implies (a-d)(a+d)=a+157}$

$\tt{\implies a^2-d^2=a+157}$

Putting the value of a=14 here]

$\tt{\implies 14^2-d^2=14+157}$

$\tt{\implies 196-d^2=171}$

$\tt{\implies -d^2=171-196}$

$\tt{\implies -d^2=-25}$

$\tt{\implies d=\sqrt{25}=5}$

Now, calculate the 10th terms of ap]

$\tt{\implies T_{n}=a+(n-1)d}$

$\tt{\implies T_{10}=14+(10-1)5}$

$\tt{\implies T_{10}=14+45}$

$\tt{\implies T_{10}=59}$

$\sf\large{Hence,}$

$\bf{\implies The\:10th\: terms\:of\:AP=59}$

$\sf{\underbrace{Answer-59}}$