Math, asked by swarajbarwal123, 1 year ago

the sum of the first seven terms of an ap is 182 if its 4th and the 17th term are in the ratio 1 : 5 find the AP

Answers

Answered by siddhartharao77

Given that the sum of the first seven terms of an AP is 182.

We know that sum of n terms of an AP Tn = n/2(2a + (n - 1) * d).

Therefore the sum of 7 terms of an AP T7 = 7/2(2a + (7 - 1) * d)

182 = 7/2(2a + 6d)

26 = 1/2(2a + 6d)

52 = (2a + 6d)

52 = 2(a + 3d)

26 = a + 3d -------- (1)

Given that 4th term and 17th term are in the ratio 1:5

T4/T17 = 1/5

(a + (4 - 1) * d)/(a + 17 - 1) * d) = 1/5

(a + 3d)/(a + 16d) = 1/5

On cross multiplication, we get

5a + 15d = a + 16d

4a = d ------ (2)

Substitute (1) in (2), we get

a + 3d = 26

a + 3(4a) = 26

a + 12a = 26

13a = 26

a = 26/13

a = 2.

Substitute a = 2 in (2),we get

4a = d

4(2) = d

8 = d.

Therefore the series is 2,10....

Hope this helps!

Answered by shreyaschavhan4

Note:-i am solving this by supposing you have good basics
Given:-
Sn=182
a4/a17=1/5_____(1)
therefore;
an=a+(n-1)d
therefore from (1)
a+3d/a+16d=1/5
5(a+3d)=a+16d
therefore 4a=d______(2)
therefore
S7=7/2(2a+6d)
182=7/2(2a+24a) ___(from 2)
182=7/2[2(a+12a)
26=13a
therefore a=2 and from (2)
d=8
and then find the terms

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