Question

13) Obtain the sum of 56 terms of an A.P. whose
19th and 38th terms are 52 and 148 respectively.​

Answer 1

$\red{\textbf{Answer}}$:-

Step-by-step explanation:

Let a and d be the first term and the common difference of the given AP respectively. Then,

a19 =52

⇒a+18d =52 ...(1)

a 38 =148

⇒a+37d =148 ...(2)

On subtracting (1) from (2), we get

19d= 96⇒

$d = \frac{96}{19}$

$putting \: d = \frac{96}{19} \: in \: (1) \: we \: get...$

$a = 52 - 18 \times \frac{96}{19}$

$= \frac{ - 740}{19}$

Now,

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

therefore,

$s_{56} = 28[ \frac{3800}{19} ]$

= 5600