
Question: A series is such that its every even term is 'a' times the term before it and every odd term is c times the term before it. The sum of 2n term of the series is (the first term is unity)
A) \[\dfrac{{(1 - {c^n})(1 - {a^n})}}{{1 - ac}}\]
B) \[\dfrac{{(1 + a)(1 - {c^n}{a^n})}}{{1 - ac}}\]
C) \[\dfrac{{(1 + {c^n})(1 + {a^n})}}{{1 - ac}}\]
D) \[\dfrac{{(1 + a)(1 + {c^n}{a^n})}}{{1 + ac}}\]
Answer
163.8k+ views
Hint: in this question we have to find sum of2n term of given series. First arrange the given series in GP once we get the series, just apply the formula of sum of series to get required value.
Formula Used: We can find sum of n terms of GP by using
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
Complete step by step solution: Given: A series is such that it’s every even term is 'a' times the term before it and every odd term is c times the term before it. First term is unity
According to question we have
\[1 + a + ca + a(ca) + c(aca) + a(caca) + ............ + 2n\]
\[1 + a + ca + {a^2}c + {a^2}{c^2} + ............ + 2n\]
\[(1 + ca + {a^2}{c^2} + ............ + n) + (a + ac + {c^2}{a^3} + ............ + nterms)\]
Now both part of above series are in GP
In order to find required value apply formula of sum of GP series.
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
\[\dfrac{{1[1 - {{(ca)}^n}]}}{{1 - ca}} + \dfrac{{a[1 - {{(ca)}^n}]}}{{1 - ca}}\]
\[\dfrac{{(1 + a)(1 - {c^n}{a^n})}}{{1 - ac}}\]
Option ‘B’ is correct
Note: Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio therefore always try to find first term and common ratio if required. Then apply the formula to get the required value.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio
Formula Used: We can find sum of n terms of GP by using
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
Complete step by step solution: Given: A series is such that it’s every even term is 'a' times the term before it and every odd term is c times the term before it. First term is unity
According to question we have
\[1 + a + ca + a(ca) + c(aca) + a(caca) + ............ + 2n\]
\[1 + a + ca + {a^2}c + {a^2}{c^2} + ............ + 2n\]
\[(1 + ca + {a^2}{c^2} + ............ + n) + (a + ac + {c^2}{a^3} + ............ + nterms)\]
Now both part of above series are in GP
In order to find required value apply formula of sum of GP series.
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
\[\dfrac{{1[1 - {{(ca)}^n}]}}{{1 - ca}} + \dfrac{{a[1 - {{(ca)}^n}]}}{{1 - ca}}\]
\[\dfrac{{(1 + a)(1 - {c^n}{a^n})}}{{1 - ac}}\]
Option ‘B’ is correct
Note: Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio therefore always try to find first term and common ratio if required. Then apply the formula to get the required value.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio
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