Question

if the sum of first 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of n terms

Answer 1

here is you answer

below

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Given,

S7 = 49

S17 = 289

$\tt{\rightarrow S7=\dfrac{7}{2}[2a+(n-1)d]}$

$\tt{\rightarrow S7=\dfrac{7}{2}[2a+(7-1)d]}$

$\tt{\rightarrow 49=\dfrac{7}{2}[2a+16d]}$

7 = (a + 3d)

a + 3d = 7 ... (1)

Now we have,

$\tt{\rightarrow S17=\dfrac{17}{2}[2a+(n-1)d]}$

$\tt{\rightarrow S17=\dfrac{17}{2}[2a+(17-1)d]}$

$\tt{\rightarrow 289=\dfrac{17}{2}[2a+16d]}$

17 = (a + 8d)

a + 8d = 17 ... (2)

Now,

Subtracting (1) from equation (2),

5d = 10

d = 2

Putting value of d in (1)

a + 3(2) = 7

a + 6 = 7

a = 1

$\tt{\rightarrow S_{n} =\dfrac{n}{2}[2a+(n-1)d]}$

$\tt{\rightarrow S_{n} =\dfrac{n}{2}[2(1)+(n-1)2]}$

$\tt{\rightarrow S_{n} =\dfrac{n}{2}[2+2n-2]}$

$\tt{\rightarrow S_{n} =\dfrac{n}{2}[2n]}$

= n²