Question

2. a) The first three terms of a geometric sequece are y, y +5 and y +15. All the terms are positive. Calculate the 5th term. b) The sum of the first three terms of a geometric series is 77 and the sum of the first six terms is 693. Find the common ratio and the first term. [8 marks]

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$t o\: find : \\ 5th \: term \: of \: a \: GP \\ \\ so \: here \\ y \: , \: y + 5 \: ,y + 15 \: \: forms \: a \: GP, \\ thus \: then \\ \\ \frac{y + 5}{y} = \frac{y + 15}{y + 5} \\ \\ applying \: dividendo \: on \: both \: sides \\ we \: get \\ \\ \frac{y + 5 - y}{y} = \frac{y + 15 - (y + 5)}{y + 5} \\ \\ \frac{5}{y } = \frac{10}{y + 5} \\ \\ \frac{1}{y} = \frac{2}{y + 5} \\ \\ y + 5 = 2y \\ \\ y = 5 \\ \\ so \: then \\ t1 = y = 5 \\ t2 = y + 5 = \: (5 + 5 \: ) = 10 \\ t 3= y + 15 =( 5 + 15 )= 20$

$so \: thus \: then \\ since \: here \: given \: sequence \: is \: a \: GP \\ a = 5 \\ r = 2 \\ \\ we \: know \: that \\ nth \: term \: of \: GP \: is \: given \: by \\ tn = a.r {}^{n - 1} \\ \\ t5= 5.(2) {}^{5 - 1} \\ \\ = 5.(2) {}^{4} \\ \\ = 5 \times 2 {}^{2} \times 2 {}^{2} \\ \\ = 5 \times 4 \times 4 \\ \\ = 16 \times 5 \\ \\ t5 = 80$

Answer 2

Step-by-step explanation:

Given: a) The first three terms of a geometric sequece are y, y +5 and y +15. All the terms are positive.

To find: Calculate the 5th term.

Solution:

nth term of GP

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

Here,

Step 1: Find value of common ratio

$\frac{y + 5}{y } = \frac{y + 15}{y + 5}$

cross multiply

$y(y + 15) = (y + 5)(y + 5)$

${y}^{2} + 15y = {y}^{2} + 10y + 25$

$15y - 10y = 25$

$5y = 25$

$\bf y = 5$

Common ratio:

$r = \frac{y + 5}{y }$

$r = \frac{10}{5}$

$\bf \blue{r =2}$

Step 2: Find 5th term

$a_5 = 5 \times (2)^{5 - 1}$

$a_5 = 5 \times {2}^{4}$

or

$\bf \pink{a_5 = 80}$

Given: b) The sum of the first three terms of a geometric series is 77 and the sum of the first six terms is 693.

To find : Find the common ratio and the first term.

Solution:

Sum of n terms of GP

$\boxed{\bf S_n = \frac{a( { {r}^{n} - 1)} }{r - 1}}, \: r > 1$

Step 1: Write equations from the given conditions

The sum of the first three terms of a geometric series is 77.

So,

$S_3 = \frac{a( { {r}^{3} - 1)} }{r - 1}$

$77= \frac{a( { {r}^{3} - 1)} }{r - 1}...eq1$

The sum of the first six terms is 693

$S_6 = \frac{a( { {r}^{6} - 1)} }{r - 1}$

$693 = \frac{a( { {r}^{6} - 1)} }{r - 1}$

or

$693 = \frac{a( { ({r}^{3})^{2} - {1}^{2} )} }{r - 1}$

or

$693 = \frac{a( { {r}^{3} - {1} )}( {r}^{3} + 1) }{r - 1}$

Place value from equation 1

or

$693 = 77( { {r}^{3} + 1})$

$( r ^{3} + 1) = \frac{693}{77}$

or

$( r ^{3} + 1) = 9$

$r ^{3} = 8$

or

${r}^{3} = {2}^{3}$

or

$\bf \red{r = 2}$

Find first term

$77= \frac{a( { {2}^{3} - 1)} }{2 - 1}$

or

$77= \frac{a( { 7)} }{1}$

$a = \frac{77}{7}$

$\bf \green{ a = 11}$

Final answer:

a) Fifth term of GP is 80.

b) First term is 11 and common ratio is 2.

Hope it helps you.

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