Question

The product of 5 terms of a GP is

243. The 3rd term of GPis equal
to 10th term of AP.Find sum of
19th term in Arthematic series.

Answer 1

The 19th term of Arithmetic series is - 2.02

Explanation:

Given as :

The product of 5 terms of a GP = 243

The 3rd term of GP is = 10th term of AP

Let The sum of 19th term of AP = x

Now, According to question

∵ nth term of GP is given by

`g_n` = a×
`r^(n - 1)`

where a is the first term and n is nth term and r is the common ratio

So For n = 1

`g_1` = a×
`r^(1 - 1)`

Or,
`g_n` = a×
`r^(0)`

i.e
`g_n` = a × 1

Or,
`g_n` = a

For n = 2

`g_2` = a×
`r^(2 - 1)`

Or,
`g_n` = a×
`r^(1)`

i.e
`g_n` = a × r

For n =3

`g_3` = a×
`r^(3 - 1)`

Or,
`g_3` = a×
`r^(2)`

i.e
`g_3` = a × r²

For n =4

`g_4` = a×
`r^(4 - 1)`

Or,
`g_4` = a×
`r^(3)`

i.e
`g_4` = a × r³

For n =5

`g_5` = a×
`r^(5 - 1)`

Or,
`g_5` = a×
`r^(4)`

i.e
`g_5` = a ×
`r^(4)`

Now, According to question

product of 5 terms of a GP = 243

So,
`g_1` ×
`g_2` ×
`g_3` ×
`g_4` ×
`g_5` = 243

Or, a × a r × a r²× a r³× a
`r^(4)` = 243

Or,
`a^(5)` ×
`r^(10)` =
`3^(5)`

∴, a r² = 3 ............1

Again

3rd term of GP = 10th term of AP

∵ nth term for AP

Tn = a + (n - 1) r , where a is first term and r is common difference

So, 10th term of AP

`A_10` = a + (10 - 1) r

`A_10` = a + 9 r

∵
`g_3` =
`A_10`

Or, a r² = a + 9 r

Now, from eq 1

3 = a + 9 r

i.e a + 9 r = 3

Or,
`(3)/(r^(2) )` + 9 r = 3

Or, 3 + 9 r³ = 3 r²

Or, 9 r³ - 3 r² + 3 = 0

Or, r = - 0.59

and a =
`(3)/((-0.59)^(2) )`

i.e a = 8.6

Now, 19th term of AP

`A_19` = a + (19 - 1) r

`A_19` = 8.6 + 18 × (-.59)

∴
`A_19` = - 2.02

So, The 19th term of Arithmetic series = tex]A_19[/tex] = - 2.02

Hence, The 19th term of Arithmetic series is - 2.02 Answer