Question

in an ap it is given that T8 =31 and T15=45 find ap

Answer 1

Answer:

2,5,8,11

Step-by-step explanation:

nth term of an AP is given by tn = a + (n – 1)d Given t8 = 23 and t17 = 9 + t14 → (1) Put n = 8 in tn, we get t8 = a + 7d ⇒ a+ 7d = 23 → (2) Put n = 17, t17 = a + 16d Put n = 14, t14 = a + 13d Hence (1), becomes a + 16d = 9 + a + 13d ⇒ 3d = 9 ∴ d = 3 Put d = 3 in equation (2), we get a + 7(3) = 23 ⇒ a = 23 – 21 = 2 Hence the AP is 2, 5, 8, 11

Answer 2

Answer:

heya mate there's your answer

let first term of the AP be T1 and common difference of the AP be d

given T8=31 and T15=45

nth term of an AP is,

Tn=T1+(n-1)d

T8=T1+(8-1)(d)

31=T1+7d______________(1)

and T15=T1+(15-1)d

45=T1+14d_____________(2)

now (1)×2 , 62=2T1+14d______(3)

solve (2)&(3)

62=2T1+14d

45=T1+14d

we get, T1=17

now apply T1=17 in (1)

31=T1+7d

31=17+7d

7d=31-17

7d=14

d=14/7

d=2

So AP is 17,19,21,23,25,27,29,31,33,.......................

hope it helps you

Thank you