Find the sum of 10 terms of an AP whose 10th term is 52 and 17th term exceeds its 13th term by 20
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Step-by-step explanation:
Given the 10'th term of an AP is 52 and its 17'th term is 20 more than its 13'th term
We have to find out,
The AP and its 30'th term
10'th term is 52 i.e. t₁₀ = a + 9d
a + 9d = 52 (given) -----[1]
its 17th term is 20 more than its 13th term
t₁₇ = 20 + t₁₃
⇒ a + 16d = 20 + a + 12d
⇒ 4d - 20 = 0
⇒ 4d = 20
⇒ d = 5
Substitute 'd' in [1]
We get,
⇒ a + 9d = 52
⇒ a + 45 = 52
⇒ a = 7
30'th term = a + 29d
7 + (29 * 5) = 152
Required answer:
AP = 7 , 12 , 17 , 22 , 27
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