Question

For what value of n ,are the nth terms if two APs 63, 65, 67... and 3, 10, 17..equal ? ​

Answer 1

Answer:

Given

(3x−5y)(9x2+25y2+15xy)

We shall use the identity

(a-b)(a2+ab+b2)=a3-b3

We can rearrange the

(35-5y)(92x2+25y2+15xy)as

=((3x-5y)((3x)2+(5y)2 +(3x)(5y))

=(3x)3-(5y)3

=(3x)×(3x)×(3x)−(5y)×(5y)×(5y)

=27x3−125y3

Hence the Product value of

(3x-5y)(92x2+25y3+15xy)is

27x3-125y3.

Answer 2

Given :

• AP are 63, 65, 67 and 3, 10, 17...

To Find :

• Value of n

Solution :

• 63, 65, 67....

$\implies \sf \: a = 63 \\ \\ </p><p> \implies \sf \: d = 65 - 63 = 2 \\ \\ \sf \implies a_{n} = a + (n - 1) d \ \\ \\ \sf\implies a_{n} = 63 + (n - 1)2$

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• 3, 10, 17.....

$\implies \sf a = 3 \\ \\ </p><p> \implies \sf d = 10 - 3 = 7 \\ \\ \implies \sf \: a_{n} = a + (n - 1) d \\ \\ \implies\sf a_{n} = 3 + (n - 1) 7 \:$

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Now, Finding value of n

$\implies \sf 63 + (n - 1) 2= 3 + (n - 1) 7\\ \\ \implies \sf63 + 2n - 2 = 3 + 7n - 7 </p><p> \\ \\ \implies \sf61 - 3 + 7 = 7n - 2n </p><p> \\ \\ \implies \sf65 = 5n </p><p> \\ \\ \implies \sf n = 65/5 </p><p> \\ \\ \implies \sf { \boxed{ \sf{ \red{n = 13}}}} \: \blue \star$

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Therefore, Value of n is 13.