In an A.P. a = – 2, d = 4 and \[{{S}_{n}}=160\]. 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 \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] 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 \[{{S}_{n}}=160\]. 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 \[{{a}_{n}}=a+\left( n-1 \right)d\] 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: \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] 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 \[{{S}_{n}}=160\]. We know that the sum of the n terms of A.P is \[{{S}_{n}}\] where \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]. By substituting the value of a = – 2, d = 4 and \[{{S}_{n}}=160\] in the above equation, we get, \[160=\dfrac{n}{2}\left[ 2\left( -2 \right)+\left( n-1 \right)4 \right]\] By multiplying 2 on both the sides of the above equation, we get, \[320=n\left( -4+4n-4 \right)\] \[\Rightarrow 320=n\left( 4n-8 \right)\] By taking out 4 common from both sides of the above equation, we get, \[4\left( 80 \right)=4n\left( n-2 \right)\] By canceling 4 from both sides of the above equation, we get, \[80=n\left( n-2 \right)\] \[\Rightarrow {{n}^{2}}-2n=80\] By subtracting 80 from both sides of the above equation, we get, \[{{n}^{2}}-2n=80\] We can also write the above equation as, \[{{n}^{2}}-10n-8n-80=0\] \[\Rightarrow n\left( n-10 \right)+8\left( n-10 \right)=0\] By taking out (n – 10) common from the above equation, we get, \[\left( n-10 \right)\left( n+8 \right)=0\] 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.
Note: Here, students must note that the number of terms can never be negative in any sequence. Also, we must understand the physical significance of A.P or arithmetic progression as well because we observe such sequence in the practical world also like roll numbers of students in class or if we have equal increments in salary, we can find the final salary easily by using the sum of the terms of A.P, etc.