Question

solve the 54nd question

Answer 1

Step-by-step explanation:

the nth term of an AP is a+(n-1)d

sum =n/2[2a+(n-1)d]

Answer 2

Given:

The sum of the first , third and seventeen term of an A.P. = 216

a + (a + 2d )+ (a + 16d) = 216

⇒3a + 18d = 216

⇒a + 6d = 72.........(i)

$S_{13} = \frac{n}{2} [2a + (n - 1)d] \\ \: \: \: \: \: \: \: = \frac{13}{2} [2a + (13 - 1)d] \\ \: \: \: \: \: \: \: = \frac{13}{2} (2a + 12d) \\ \: \: \: \: \: \: \: = \frac{13}{2} \times 2(a + 6d) \\ \: \: \: \: \: \: \: = 13 \times 72 = 936$

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