if the 10th term of an Ap is52 and 17th term is 20more than 13th term find Ap
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3
hello....
given, a10 =52
a + 9d =52 ...........(1)
also, a17 = 20+a13
a + 16d =20+a+12d
16d-12d=20
4d=20
d = 5
putting the value of d in eq. (1), we get,
a + 9(5) =52
a +45 = 52
a = 7
hence, the required AP is 7,12,17,.........
I hope it help you...
please mark it as a brainieist answer
given, a10 =52
a + 9d =52 ...........(1)
also, a17 = 20+a13
a + 16d =20+a+12d
16d-12d=20
4d=20
d = 5
putting the value of d in eq. (1), we get,
a + 9(5) =52
a +45 = 52
a = 7
hence, the required AP is 7,12,17,.........
I hope it help you...
please mark it as a brainieist answer
Answered by
2
Answer:
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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