Question

the second term of a GP is 2 and the sum of infinite terms in GP is 8 find the first term

Answer 1

a = first term and r = common ratio
ar = 2  -   equation 1    and     a/1-r= 8  - equation 2
so r = 2/a from equation 1 substituting it in equation 2
a = -8r+8
a = -8*2/a +8
a + 16/a = 8
$a^{2}$ + 16 = 8a
$a^{2}$ - 8a +16 =0
$a^{2}$ - 4a - 4a +16
a(a-4)-4(a-4)
a-4 = 0
so a = 4 and the first term = a

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is 4

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$\huge\sf\blue{Given}$

✭ First term of a GP = 2

✭ Sum of infinite terms in GP is 8

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$\huge\sf\gray{To \:Find}$

◈ The first term?

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$\huge\sf\purple{Steps}$

≫ Let the first term be a and common ratio be r.

$\dashrightarrow \sf Given \: \: T_2 = 2 \\\\ \sf\dashrightarrow ar =2 \qquad -eq(1)\\\sf Also \: Given \\\\ \sf\dashrightarrow S \infty = 8 \\\\ \sf\dashrightarrow\frac{a}{1 - r} = 8 \\\\\sf\dashrightarrow a = 8 - 8r\\ \\ \sf\dashrightarrow substituting \: \: from \: \: (1) \\\\ \sf\dashrightarrow\frac{2}{r} = 8 - 8r \\\\\sf\dashrightarrow 4 {r}^{2} - 4r + 1 = 0 \\\\ \sf\dashrightarrow {(2r - 1)}^{2} = 0 \\\\ \sf\dashrightarrow r = \frac{1}{2} \\\\ \sf\dashrightarrow a = \frac{2}{r} = 4 \\ \\ \sf \therefore \: \: GP \: \: is 4,2,1, \frac{1}{2} ......$

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