Question

The 4th and 7th and the last term of GP are 10 ,80 and 2560 is respectively find the first term and number of terms in GP

Answer 1

Answer:

First term

$a = \frac{5}{4}$

number of terms n= 12

Step-by-step explanation:

To find the first term of GP if 4th and 7th and the last term of GP are 10 ,80 and 2560 is respectively

nth term of GP

$a_n= a {r}^{n - 1}$

a: first term

r: common ratio

4th term

$a_4= a {r}^{3} = 10...eq1$

$a_7 = a {r}^{6} = 80 ...eq2$

$a_n = a {r}^{n - 1} = 2560...eq3$

Divide eq2 by eq1

$\frac{a {r}^{6} }{a {r}^{3} } = \frac{80}{10} \\ \\ {r}^{3} = 8 \\ \\ {r}^{3} = {(2)}^{3} \\ \\ r = 2 \\ \\ \because \: when \: powers \: are \: same \: base \: can \: be \: compared$

To find the first term:

$a {r}^{3} = 10 \\ \\ a( {2)}^{3} = 10 \\ \\ a = \frac{10}{8} \\ \\ a = \frac{5}{4}$

To find total number in GP:

$2560 = a {r}^{n - 1} \\ \\ 2560 = \frac{5}{4} {(2}^{n - 1)} \\ \\ \frac{2560 \times 4}{5} = {2}^{n - 1} \\ \\ 512 \times 4 = {2}^{n - 1} \\ \\ {2}^{9} {2}^{2} = {2}^{n - 1} \\ \\ {2}^{11} = {2}^{n - 1} \\ \\ compare \: powers \\ \\ n - 1 = 11 \\ \\ n = 12$

Hope it helps you.