Question

The sum of first 16 terms of an ap 112 and sum of its next fourteen term is 518 .find ap

Answer 1

S16 =
$\frac{n}{2} (2a + (n - 1)d) \\ 112 = \frac{16}{2} (2a + 15d) \\ 112 = 8(2a + 15d) \\ 112 = 16a + 120d \\ 8a + 60d = 56 \\ 4a + 30d = 28 \\ 2a + 15d = 14......1$
there after the sum of 14 terms is 518.
there are 16+14=30 terms.
therefore the sum of 30 terms is 112+518=630.
S30=
$518= \frac{30}{2} (2a + 29d) \\ 518 = 15(2a + 29d)518 \\ 518 = 30a + 435d \\ 30a + 435d = 518....2$
multiply 1 with -2
we get
$- 30a - 225d = - 210.....3$
adding 2 and 3 we get
$210d = 308 \\ d = \frac{308}{210} \\ d = \frac{22}{15}$
put d=22/15 in 1
$2a + 15 \times \frac{22}{15} = 14 \\ 2a = 14 - 22 \\ 2a = - 8 \\ a = - 4$
therefore the A.P is
$- \frac {38}{15} - \frac {16}{15} \frac{2}{5} \frac{28}{15} .......$
hope this helps you:-)))))
don't forget to do it brainleist answer.

Answer 2

Step-by-step explanation:

Sum of first 6 terms is 112

Sum of next 14 terms is 518

Then,sum of 20terms =sum of first 6terms and sum of next 14terms

112+518=630=S20

S6=6/2(2a+5d)

112×2/6=2a+5d

112/3=2a+5d.......(1)

S20=20/2(2a+19d)

630/10=2a+19d.....(2)

Subtracting eq(1)byeq(2)

37.3=2a+5d

63=2a+19d

= - 25.7= - 14d

d=25.7/14.10/10

257/140=1.8

d=1.8