Question

16. The sum of first 10 terms of an A.P is 100 and the sum of its first 9 terms is 81, then find the 10 th term of the same A.P​

Answer 1

Answer:

If 100 is the sum of first ten terms of an AP and if 81 is the sum of first 9 terms then the tenth term is 19

Step-by-step explanation:

Given

Sum of first terms 10 of AP = 100

Given the sum of its first 9 terms = 81

So the tenth number = Sum of first terms 10 of AP -the sum of its first 9 terms

= 100-81

= 19

So the tenth number is 19

Answer 2

Answer:

10th term of the AP is 169

Step-by-step explanation:

S9=81

$\\ \frac{9}{2}[2a+(9-1)d]=81$

S10=100

$\\ \frac{10}{2} [2a+(10-1)d]=100$

adding S9 and S10

$\\ 9[2a+8d]=162\\18a+72d=162$           (eqn 1)

$5[2a+9d]=100\\10a+45d=100$             (eqn 2)

taking 5 common from eqn 2

$2a+9d=50$                  (eqn 3)

now subtracting eqn 3 from eqn 1

$\\ 18a+72d=162\\9[2a+9d=50]\\\\18a+81d=450\\ 18a+72d=162$

we get,

$\\ 9d=288\\d=32$

now substituting d=32 in eqn 3

$2a+9(32)=50\\2a+288=50\\2a=-238\\a=-119$

a= -119, d=32

an=$a+(n-1)d$

a10= -119+(10-1)32

a10= -119+9*32

a10= -119+288

a10= 169