Question

Find five consecutive terms in an A.P such that their sum is 60 and the product of the third and the fourth term exceeds the fifth by 172.

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Answer 1

GIVEN :-

sn = 60

n = 5

4tt term × 3rd term = 5th term + 172

TO FIND :-

we have to find ap

SOLUTION :-

let the ap be :-

a - 2d , a - d , a , a + d , a + 2d

[ note :- these are called ready made ap ]

now according to question :-

sum of Ap = 60

=> (a - 2d) + (a - d) + (a) + (a + d ) + (a + 2d) = 60

=> 5 a = 60

=> a = 12

now putting the condition given in question :-

=> 4th term × 3rd term = 5th term + 172

=> (a + d) (a) = (a + 2d) + 172

=> a² + ad = a + 2d + 172

now put the value of a = 12

=> (12)² + (12) d = 12 + 2d + 172

=> 144 + 12d = 184 + 2d

=> 10d = 40

=> d = 4

HENCE , ap =

=> 4 , 8 , 12 , 16 , 20

Answer 2

Step-by-step explanation:

Let the terms of AP be

A−2d,a−d,a,a+d,a+2d

A/Q

(a−2d)+(a−d)+a+a(a+d)+(a+2d)=60

5a=60⇒a=12

and a×(a+d)=172+a+2d

⇒a

2

+ad=172+a+2d

⇒144+12d=172+12+2d

⇒12d−2d=184−144=40

⇒10d=40⇒d=4

∴a=12,d=4

and the terms are

12−4×2,12−4,12,12+4,12+2×4

=4,8,12,16,20