Question

Find the 20th term of an AP, whose 5th term is 17 and 9th term is exceeds 2nd term by 35. And also find sum of its 20 term

Answer 1

Answer:

t₂₀ = 92 .

S₂₀ = 890

Step-by-step explanation:

Given :

t₅ = 17 ... ( i )

t₉ = t₂ + 35 ... ( ii )

Let first term be a and common difference d.

We know :

t _ n = a + ( n - 1 ) d

From ( i )

17 = a + 4 d

a = 17 - 4 d ... ( iii )

From ( ii )

a + 8 d = a + d + 35

7 d = 35

d = 5

Putting value of d in ( iii )

a = 17 - 20

a = - 3

Now ,

t₂₀ = - 3 + 19 × 5

t₂₀ = 92 .

We know sum formula :

S_n = n / 2 [ 2 a + ( n - 1 ) d ]

S₂₀ = 20 / 2 [ 2 × - 3 + 19 × 5 ]

S₂₀ = 10 ( - 6 + 95 )

S₂₀ = 890