The 5th term of an arthmetic sequence is 40 and 10th term 20 . Find 15th term , how many terms of this sequence make the sum zero
Answers
Answer:
15th term = 0
sum of 29 terms = 0
Step-by-step explanation:
5th term = a + 4d = 40
10th term = a + 9d = 20
10th - 5th
(a + 9d) - (a + 4d) = 20 - 40
a + 9d - a - 4d = -20
5d = - 20
d = -20/5 = -4
substitute d value in
a + 4d = 40
a + 4(-4) = 40
a - 16 = 40
a = 56
15th term is
a + 14d = 56 + 14(-4) = 56 - 56 = 0
\begin{gathered}sum \: = 0 \\ \frac{n}{2} (2a + (n - 1)d) = 0 \\ \frac{n}{2} (2(56) + (n - 1) \times ( - 4)) \\ \frac{n}{2} (112 - 4n + 4) \\ \frac{n}{2} (116 - 4n) \\ \frac{n}{2} \times 2(58 - 2n) \\ 58n - 2 {n}^{2} = 0 \\ 2n(29 - n) = 0 \\ 2n \: = 0 \: \: \: \: 29 - n = 0 \\ we \: take \: 29 - n = 0 \\ n = 29 \\ therefore \: sum \: of \: 29 \: terms \: makes \: 0\end{gathered}
sum=0
2
n
(2a+(n−1)d)=0
2
n
(2(56)+(n−1)×(−4))
2
n
(112−4n+4)
2
n
(116−4n)
2
n
×2(58−2n)
58n−2n
2
=0
2n(29−n)=0
2n=029−n=0
wetake29−n=0
n=29
thereforesumof29termsmakes0
Step-by-step explanation:
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4d= -20. a= 60
d=-5
Fifteenth terms = 60+14(-5)= -10