in an ap if a -18 n 10 an 0 find d
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CBSE
Mathematics
Grade 10
Arithmetic Progression
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In an A.P. a = – 2, d = 4 and Sn=160
Sn=160
. Find the value of n.
Answer
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Hint: First of all try to recollect what an A.P is. Now use the formula for the sum of the terms in A.P that is Sn=n2[2a+(n−1)d]
Sn=n2[2a+(n−1)d]
and substitute in it the given values to get the value of n.
Complete step-by-step answer:
Here, we are given an A.P in which a = – 2, d = 4 and Sn=160
Sn=160
. We have to find the value of n. Before proceeding with the question, let us see what an A.P is. A.P or Arithmetic Progression is a sequence of numbers so that the difference between two successive numbers is a constant value. For example, the series of odd numbers is 1, 3, 5, 7….. is an A.P, which has a common difference of 2 between successive terms. We can also write the general term of an A.P is an=a+(n−1)d
an=a+(n−1)d
where ‘a’ is the first term, ‘d’ is the common difference and n is the number of the terms in A.P.
Also, the sum of the n terms of an A.P is given as:
Sn=n2[2a+(n−1)d]
Sn=n2[2a+(n−1)d]
where a, d and n have their usual meanings. Now, let us consider our question. Here we are given an A.P in which the first term that is a = – 2, common difference that is d = 4 and sum of the n terms that is Sn=160
Sn=160
.
We know that the sum of the n terms of A.P is Sn
Sn
where Sn=n2[2a+(n−1)d]
Sn=n2[2a+(n−1)d]
.
By substituting the value of a = – 2, d = 4 and Sn=160
Sn=160
in the above equation, we get,
160=n2[2(−2)+(n−1)4]
160=n2[2(−2)+(n−1)4]
By multiplying 2 on both the sides of the above equation, we get,
320=n(−4+4n−4)
320=n(−4+4n−4)
⇒320=n(4n−8)
⇒320=n(4n−8)
By taking out 4 common from both sides of the above equation, we get,
4(80)=4n(n−2)
4(80)=4n(n−2)
By canceling 4 from both sides of the above equation, we get,
80=n(n−2)
80=n(n−2)
⇒n2−2n=80
⇒n2−2n=80
By subtracting 80 from both sides of the above equation, we get,
n2−2n=80
n2−2n=80
We can also write the above equation as,
n2−10n−8n−80=0
n2−10n−8n−80=0
⇒n(n−10)+8(n−10)=0
⇒n(n−10)+8(n−10)=0
By taking out (n – 10) common from the above equation, we get,
(n−10)(n+8)=0
(n−10)(n+8)=0
So, we get n = 10, n = – 8.
As we know the number of terms could not be negative. So, we get the number of terms in A.P as 10.