The sum of 5th and 7th term of an ap is 52 and the 10th term is 46 find the
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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,...
Step-by-step explanation:
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