Question

Prove that the square of any term of the arithmetic sequence 7,11,15.....will not be a term of the sequence

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given:Prove that the square of any term of the arithmetic sequence 7,11,15.....will not be a term of the sequence.

To find: Proof of square of any term will not be the tem of given sequence

Solution:

In given A.P.

7,11,15...

First term a= 7

Common difference d= 4

nth term can be written as

$\bold{\red{a_n = a + (n - 1)d}} \\ \\ a_n = 7 + (n - 1)4 \\ \\ a_n = 7 - 4 + 4n \\ \\ \bold{\green{a_n = 3 + 4n}}$

Now take any term,let n= 8

$a_8 = 3 + 4 \times 8 \\ \\ a_8 = 3 + 32 \\ \\ a_8 = 35$

Now take square of 35

$( {a_8)}^{2} = ( {35)}^{2} = 1225$

check,whether 1225 is a term or not

$1225 = 3 + 4n \\ \\ 1225 - 3 = 4n \\ \\ 4n = 1222 \\ \\ n = \frac{1222}{4} \\ \\ n = 305.5$

n is rational(have decimal point)

but ,

we know that number of terms should not be rational.

number of terms are positive integers.

Thus 1225 will not be the any term of given sequence/A.P.

By this way,this can be proved by taking another's terms too.

In general:

Let square of an

$( {a_n)}^{2} = ( {3 + 4n)}^{2} \\ \\( {a_n)}^{2} = 9 + 16 {n}^{2} + 24n$

if this is the term of A.P.

$3 + 4n = 9 + 16 {n}^{2} + 24n \\ \\ 16 {n}^{2} + 20n + 6 = 0 \\ \\ 8 {n}^{2} + 10n + 3 = 0 \\ \\ n_{1,2}= \frac{ - 10 ± \sqrt{100 - 96} }{16} \\ \\ n_{1,2} = \frac{ - 10 ±2}{16} \\ \\ n_1 = \frac{ - 8}{16} = \frac{ - 1}{2} \neq \: integer \\ \\ n_2 = \frac{ - 12}{16} = \frac{ - 3}{4} \neq \: integer$

Thus,we can conclude that square of any term can not be the term of that A.P./sequence

Hope it helps you.

To learn more on brainly:

1)26th, 11th and last term of an ap are 0, 3 and -1/5 respectively . find the common difference and the number of terms

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2)If the sum of the first 8 terms of AP is 136 and that of first 15 terms is 465 then find the sum of first 25 terms

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