Question

obtain the sum of the first 56 terms of an A.P whose 18th term and 39th term is 52 and 148​

Answer 1

Given :

• 18th term of AP = 52
• 39th term of AP = 148

To find :

Sum of first 56 terms

Formula used :

$\sf \bullet \: \: T_n = a + (n-1)d$

$\sf \bullet \: \: S_n \: = \: \dfrac{n}{2} \bigg(2a \: + (n - 1)d \bigg)$

Here,

• a = 1st term
• n = number of terms
• d = common difference
• Tₙ = nth term
• Sₙ = Sum of first n terms

Solution :

$\sf \bullet \: \: S_{56} \: = \: \dfrac{56}{2} \bigg(2a \: + (56 - 1)d \bigg)$

$\sf \implies \: \: S_{56} \: = \: \dfrac{56}{2} \bigg(2a \: + 55d \bigg) \: \: \: \: \: equation1$

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$\sf \: \: T_{18} \: = a + (18 - 1)d$

$\sf \implies \: 52 \: = a + 17d \: \: \: \: \: \: \: equation \: 2$

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$\sf \bullet \: \: T_{39} \: = a + (39 - 1)d$

$\sf \implies \: 148\: = a + 38d \: \: \: \: \: \: \: equation \: 3$

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Now , add equation 2 and equation 3

148 + 52 = (a+38d) + (a+17d)

➝ 200 = a + 38d + a + 17d

➝ 2a + 55d = 200 equation 4

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Now put value of (2a+55d) from equation 4 into equation 1.

$\sf \implies \: \: S_{56} \: = \: \dfrac{56}{2} \bigg(200 \bigg)$

$\sf \implies \: \: S_{56} \: = \: 28 \times 200$

$\sf \implies \: \: S_{56} \: = \: 5600$

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ANSWER : 5600

Answer 2

Answer:

5600

Step-by-step explanation:

❤Hope it helps

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