If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.
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6
Answer:
The required A.P. is 3, 10, 17,…
Step-by-step explanation:
Given :
a5 = 31 & a25 = 140 + a5
Case 1 :
a5 = 31
By using the formula , nth term ,an = a + (n -1)d
a + (5 -1)d = 31
a + 4d = 31 ………….(1)
a = 31 - 4d …………..(2)
Case 2 :
a25 = 140 + a5
a + (25 - 1)d = 140 + (a + (5 -1)d
a + 24d = 140 + a + 4d
a + 24d = 140 + 31
[From eq 1]
(31 - 4d) + 24d = 171
[From eq 2]
-4d + 24d = 171 - 31
20d = 140
d = 140/20
d = 7
On putting the value of d = 7 in eq (2)
a = 31 - 4d
a = 31 – 4(7)
a = 31 - 28
a = 3
First term, a1 = a = 3
Second term , a2 = (a +d) = 3 + 7 = 10
Third term, a3 = (a +2d) = 3 + 2(7) = 3 + 14 = 17
Hence, the required A.P. is 3, 10, 17,…
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