Question:
6. Obtain the sum of the 56 terms of an A. P. whose 19th and 38th terms are 52 and 148 respectively.
Sol. Given: \( t_{19} = 52 \), \( t_{38} = 148 \)
We know the formula for the \( n^{th} \) term of an A.P.:
\( t_n = a + (n - 1)d \)
For the 19th term:
\( \therefore t_{19} = a + (19 - 1)d \)
\( \therefore 52 = a + 18d \)
\( \therefore a + 18d = 52 \) — (i)
\( \therefore t_{19} = a + (19 - 1)d \)
\( \therefore 52 = a + 18d \)
\( \therefore a + 18d = 52 \) — (i)
For the 38th term:
\( t_{38} = a + (38 - 1)d \)
\( \therefore 148 = a + 37d \)
\( \therefore a + 37d = 148 \) — (ii)
\( t_{38} = a + (38 - 1)d \)
\( \therefore 148 = a + 37d \)
\( \therefore a + 37d = 148 \) — (ii)
Adding equations (i) and (ii):
\( (a + 18d) + (a + 37d) = 52 + 148 \)
\( \therefore 2a + 55d = 200 \) — (iii)
\( (a + 18d) + (a + 37d) = 52 + 148 \)
\( \therefore 2a + 55d = 200 \) — (iii)
Now, we need to find the sum of the first 56 terms (\( S_{56} \)).
The sum formula is: \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( \therefore S_{56} = \frac{56}{2}[2a + (56 - 1)d] \)
\( \therefore S_{56} = 28[2a + 55d] \)
\( \therefore S_{56} = 28[2a + 55d] \)
Substituting the value from equation (iii):
\( \therefore S_{56} = 28 [200] \) [From Eq. (iii)]
\( \therefore S_{56} = 5600 \)
\( \therefore S_{56} = 28 [200] \) [From Eq. (iii)]
\( \therefore S_{56} = 5600 \)
\( \therefore \) Sum of the first 56 terms of the A.P. is 5600.