Question

Ex. (5) - The 10th term and the 18 th term of an A.P. are 25 and 41 respectively then
find 38th term of that A.P., similarly if nth term is 99. Find the value of n.​

Answer 1

Step-by-step explanation:

$\sf {T}_{10}=25 \:\:\:\: {T}_{18} = 41 \\ \\ \sf We\: know\: that \\ \\ \sf {T}_{n} = a + (n-1)d \\ \\ \sf a +9d = 25 \\ \\ \sf a + 17d = 41 \\ \\ \sf Subtracting\: above \: equation \\ \\ \sf 8d = 18 \\ \\ \sf d = 2 \\ \\ \sf a = 7 \\ \\ \sf {T}_{38} = 7 + 37 × 2 = 81$

Answer 2

Answer:

given 10th is 25

18th term is 41

let

a+9d=25

a+17d=41

subtract them

then A gets cancelled

-8d=-16

d=16÷8=2

put it in equation 1

a+9d=25

a+9×2=25

a+18=25

a=25-18

a=7.

an=a+(n-1)d=99

7+(n-1)2=99

(n-1)2=99-7

(n-1)2=92

(n-1)=92÷2

(n-1)=46

n=46+1

n=47

• so 99 is 47th term