in a geometric series of the sixth term is 16 times the second term and the sum of first seven term is 127/4,then find the positive common ratio and the first term of the series
2 thoughts on “in a geometric series of the sixth term is 16 times the second term and the sum of first seven term is 127/4,then find the positiv”
ANSWER:
Given:
6th term of a GP = 16 times the 2nd term.
Sum of first 7 term = 127/4
To Find:
Common Ratio (+ive)
First term of series.
Solution:
[tex]\text{We know that, n$_{th}$ term of a GP is,}\\\\:\longrightarrow a_{n}=ar^{n-1}\\\\\text{Here, a=1st term, r=common ratio and n=number of term.}\\\\\text{We are given that,}\\\\:\longrightarrow a_6=16\times a_2\\\\:\implies ar^{6-1}=16ar^{2-1}\\\\:\implies a\!\!\!/\:r^5=16a\!\!\!/\:r\\\\:\implies\dfrac{r^{5\!\!\!/\:\:4}}{r\!\!\!/}=16\\\\:\implies r^4=16\\\\:\implies r=\pm2\\\\\text{We are given that ratio should be positive. So,}\\\\\bf{:\implies r=+2}[/tex]
[tex]\\\text{We know that, sum of n terms of a GP is,}\\\\:\longrightarrow S_n=\dfrac{a(r^n-1)}{r-1}\\\\\text{Here, a=1st term, r=common ratio and n=number of term.}\\\\\text{We are given that,}\\\\:\longrightarrow S_7=\dfrac{127}{4}\\\\:\implies\dfrac{a(2^7-1)}{2-1}=\dfrac{127}{4}\\\\:\implies\dfrac{a(128-1)}{1}=\dfrac{127}{4}\\\\:\implies\dfrac{127\!\!\!\!\!\!/\:\:a}{1}=\dfrac{127\!\!\!\!\!\!/\:\:}{4}\\\\\bf{:\implies a=\dfrac{1}{4}}[/tex]
ANSWER:
Given:
To Find:
Solution:
[tex]\text{We know that, n$_{th}$ term of a GP is,}\\\\:\longrightarrow a_{n}=ar^{n-1}\\\\\text{Here, a=1st term, r=common ratio and n=number of term.}\\\\\text{We are given that,}\\\\:\longrightarrow a_6=16\times a_2\\\\:\implies ar^{6-1}=16ar^{2-1}\\\\:\implies a\!\!\!/\:r^5=16a\!\!\!/\:r\\\\:\implies\dfrac{r^{5\!\!\!/\:\:4}}{r\!\!\!/}=16\\\\:\implies r^4=16\\\\:\implies r=\pm2\\\\\text{We are given that ratio should be positive. So,}\\\\\bf{:\implies r=+2}[/tex]
[tex]\\\text{We know that, sum of n terms of a GP is,}\\\\:\longrightarrow S_n=\dfrac{a(r^n-1)}{r-1}\\\\\text{Here, a=1st term, r=common ratio and n=number of term.}\\\\\text{We are given that,}\\\\:\longrightarrow S_7=\dfrac{127}{4}\\\\:\implies\dfrac{a(2^7-1)}{2-1}=\dfrac{127}{4}\\\\:\implies\dfrac{a(128-1)}{1}=\dfrac{127}{4}\\\\:\implies\dfrac{127\!\!\!\!\!\!/\:\:a}{1}=\dfrac{127\!\!\!\!\!\!/\:\:}{4}\\\\\bf{:\implies a=\dfrac{1}{4}}[/tex]
Formulae Used:
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