Question

obtain the sum of first 56 terms of an ap whose 19th and 36th terms are 52 and 148 friends please help​

Answer 1

$\bf{\underline{Correct\:Question:-}}$

Note :-

⚔️ Check your question Carefully .

• obtain the sum of first 56 terms of an ap whose 19th and 38th terms are 52 and 148.

$\bf{\underline{Given:-}}$

• $\sf t_{19} = 56 , t_{38} = 148$

So,

$\sf → t_n = a+(n-1)d$

$\sf → t_{19} = a + (19-1)d$

$\sf → 52 = a + 18d$

$\bf\large→ a + 18d =52---equ(¡)$

Now,

$\sf → t_{36} = a+(n-1)d$

$\sf → t_{36} = a +(36-1)d$

$\sf → 148 = a + 35d$

$\bf\large → a+37d=148---equ(¡¡)$

• Adding equ(¡) and equ(¡¡)

$\sf → a + 18d = 52 + a + 37d = 148$

$\sf → a + 18d + a + 37d = 52 + 148$

$\large\bf → 2a + 55d = 200---equ(¡¡¡)$

$\bf{\underline{Therefore:-}}$

$\bf\large → S_n = \frac{n}{2}[2a+(n-1)d]$

$\sf → S_{56} = \frac{56}{2}[2a + (56-1)d]$

$\sf → S_{56} = 28 [ 2a + 55d]$

$\sf → S_{56} = 28[ 200]$【 from equ(¡¡¡) 】

$\sf → S_{56} = 5600$

$\bf{\underline{Hence:-}}$

• The sum of 56th term of an A.P is 5600