Question 6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
Class X1 - Maths -Sequences and Series Page 185
Answers
Answered by
2
Given,AP is 25, 22, 19.....
Here,
First term(a) = 25
common difference (d) = 22-25 = -3
sum of nth term (Sn) = 116
use, formula,
Sn = n/2{2a + (n-1)d}
116 =n/2{2×25 + (n-1)×(-3)}
232 =n{50 -3n+3}
232 =53n-3n²
3n² -53n + 232 = 0
3n² -24n -29n +232 = 0
3n(n-8) -29(n -8) = 0
(3n -29)(n-8) = 0
n = 8, 29/3
But n can't be fraction.
So, n = 8
Now, use formula for nth term
Tn = a + (n-1)d
T8 = 25 + (8-1)×(-3)
= 25 - 21
= 4
Here,
First term(a) = 25
common difference (d) = 22-25 = -3
sum of nth term (Sn) = 116
use, formula,
Sn = n/2{2a + (n-1)d}
116 =n/2{2×25 + (n-1)×(-3)}
232 =n{50 -3n+3}
232 =53n-3n²
3n² -53n + 232 = 0
3n² -24n -29n +232 = 0
3n(n-8) -29(n -8) = 0
(3n -29)(n-8) = 0
n = 8, 29/3
But n can't be fraction.
So, n = 8
Now, use formula for nth term
Tn = a + (n-1)d
T8 = 25 + (8-1)×(-3)
= 25 - 21
= 4
Answered by
1
Answer is in attachment.
Attachments:
Similar questions