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Answer 1

Answer:

Question 2:

Sum of a Geometric Progression is given by the formula:

$\boxed{S_n = \dfrac{ a(r^n - 1) }{r - 1 }}$

According to the question (2),

• Sum of 'n' terms = (19531 / 625)
• First term 'a' = 25
• Last Term 'l' = (1 / 625)

The general form of 'nth' term in a G.P. is:

$\implies{ \boxed{ a_n = a.\:r^{n-1}}$

Hence, substituting the last term, we get:

$\implies \dfrac{1}{625} = 25.\:r^{n-1}\\\\\\\implies \dfrac{1}{625} \times \dfrac{1}{25} = r^{n-1}\\\\\\\implies \dfrac{1}{15625} = r^{n-1}\\\\\\\implies \dfrac{r^n}{r} = \dfrac{1}{15625}\\\\\\\implies \boxed{ r^n = \dfrac{r}{15625} }$

Substituting the value of r to power n in Sum formula we get:

$\implies \dfrac{19531}{625} = 25 \times \dfrac{ ( \dfrac{r}{15625} - 1)}{ r - 1 }\\\\\\\implies \dfrac{19531}{625} = 25 \times \dfrac{ \dfrac{(r - 15625)}{15625}}{r-1}\\\\\\\implies \dfrac{19531}{625} = \dfrac{25(r-15625)}{15625(r-1)}\\\\\\\implies \dfrac{19531}{625} = \dfrac{ r - 15625}{625(r-1)}\\\\\\\text{Cancelling 625 from both sides we get:}\\\\\\\implies 19531 = \dfrac{r - 15625}{r-1}\\\\\\\implies 19531(r-1) = r - 15625\\\\\\\implies 19531r - 19531 = r - 15625$

$\implies 19531r - r = 19531 - 15625\\\\\\\implies 19530r = 3906\\\\\\\implies \boxed{\bf{r = \dfrac{3906}{19530} = \dfrac{1}{5}}}$

Hence Option (1) is the correct answer.

Question 3:

Given that,

$\implies a_{3} = \dfrac{27}{8}\:\: and\:\: a_{8} = \dfrac{4}{9}\\\\\\\implies \dfrac{a_8}{a_3} = \dfrac{ar^{(8-1)}}{a r^{(3-1)}}\\\\\\\implies \dfrac{a_8}{a_3} = \dfrac{ar^7}{ar^2}\\\\\\\implies \dfrac{a_8}{a_3} = r^5 \:\:\: ...(i)\\\\\\\text{Substituting the values of } a_8 \text{ and } a_3 \text{, we get:}\\\\\\\implies \dfrac{(4/9)}{(27/8)} = r^5\\\\\\\implies \dfrac{4}{9} \times \dfrac{8}{27} = r^5\\\\\\\implies \dfrac{32}{243} = r^5\\\\\\\implies (\dfrac{2}{3})^5 = r^5$

$\implies \boxed{\bf{r = \dfrac{2}{3}}}\\\\\\\implies a_{3} = a.\:r^2\\\\\\\implies \dfrac{27}{8} = a.\:\dfrac{2^2}{3^2}\\\\\\\implies a = \dfrac{27}{8} \times \dfrac{9}{4} = \dfrac{243}{32}\\\\\\\implies a = (\dfrac{3}{2})^5$

Now calculating the 12th term we get:

$\implies a_{12} = a.\;r^{12-1} = ar^{11}\\\\\\\implies a_{12} = \dfrac{3^5}{2^5} \times \dfrac{2^{11}}{3^{11}}\\\\\\\implies a_{12} = \dfrac{2^{(11-5)}}{3^{(11-5)}}\\\\\\\implies \boxed{ \bf{ a_{12} = \dfrac{2^6}{3^6}}}$

Hence Option (3) is the correct answer.

Question 4)

Given G.P = 27/8, 9/4, 3/2, ... ∞

From the given series, we can see that:

• a = 27/8
• r = (27/8) ÷ (9/4) = 2/3

Since 'r' < 1, the sum of infinite terms would be finite. Hence substituting in the Sum of an infinite G.P Formula, we get:

$\boxed{ \text{Sum of an Infinite G.P.} = \dfrac{a}{1-r}}$

$\implies S_\infty = \dfrac{27/8}{1-(2/3)}\\\\\\\implies S_\infty = \dfrac{(27/8)}{ (\dfrac{3-2}{3}) }\\\\\\\implies S_\infty = \dfrac{27/8}{1/3}\\\\\\\implies S_\infty = \dfrac{27}{8} \times \dfrac{3}{1}\\\\\\\implies \boxed{ \bf{ S_\infty = \dfrac{81}{8}}}$

Hence Option 2) is the correct answer.

Question 5)

Let us first split the given series into two different series 'A' and 'B'.

'A' = 1/5 + 1/5² + ... ∞

'B' = 1/7 + 1/7² + ... ∞

Sum of both the series = Required Answer.

Calculating Sum of 'A' we get:

• a = 1/5
• r = 1/5

$\implies S_\infty = \dfrac{a}{1-r}\\\\\\\implies S_\infty = \dfrac{1/5}{1-1/5}\\\\\\\implies S_\infty = \dfrac{1/5}{ (5-1)/5}\\\\\\\implies S_\infty = \dfrac{1/5}{4/5}\\\\\\\implies S_\infty = \dfrac{1}{5} \times \dfrac{5}{4} \\\\\\\implies \boxed{S_\infty = \dfrac{1}{4}}$

Hence Sum of 'A' = 1/4.

Calculating Sum of 'B', we get:

• a = 1/7
• r = 1/7

$\implies S_\infty = \dfrac{a}{1-r}\\\\\\\implies S_\infty = \dfrac{1/7}{1-1/7}\\\\\\\implies S_\infty = \dfrac{1/7}{ (7-1)/7}\\\\\\\implies S_\infty = \dfrac{1/7}{6/7}\\\\\\\implies S_\infty = \dfrac{1}{7} \times \dfrac{7}{6} \\\\\\\implies \boxed{S_\infty = \dfrac{1}{6}}$

Hence Sum of 'B' = 1/6

Adding both the sums, we get:

$\implies S_{G.P} = S_A+S_B\\\\\\\implies S_{G.P} = \dfrac{1}{4} + \dfrac{1}{6}\\\\\\\implies S_{G.P} = \dfrac{6+4}{24}\\\\\\\implies \boxed{ \bf{ S_{G.P} = \dfrac{10}{24} = \dfrac{5}{12}}}$

Hence the sum of the given G.P. series is 5/12.

Hence Option (1) is the correct answer.