Question

(i) The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find
A.P.
15. If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term
TA​

Answer 1

Let the first term and common difference of AP are a and d, respectively.
According to the question,
a5 + a7 = 52 and a10 = 46
⇒ a + (5-1)d + a + (7-1)d = 52 [∵an = a + (n-1)d]
and a + (10-1)d = 46
⇒ a + 4d + a + 6d = 52
and a + 9d = 46
⇒ 2a + 10d = 52
and a + 9d = 46
⇒ a + 5d = 26 ...(i)
a + 9d = 46
On subtracting Eq.(i) from Eq.(ii),we get
4d = 20 ⇒ d = 5
From Eq.(i), a = 26 - 5(5) = 1
So, required AP is a,a+d,a+2d,a+3d,...i.e., 1,1 + 5,1 + 2(5), 1 + 3(5),... i.e.,
1,6,11,16,...

Answer 2

Answer:

answer 0f (i)

Step-by-step explanation:

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