Question

which term of the AP...3,15,27,39.........will be 132 more than its 54th term????​

Answer 1

Step-by-step explanation:

I I do not understand good question

Answer 2

Answer:

3,15 , 27,39........

First we have to calculate 54th team

we know that

${a}^{n} = a + (n - 1)d$

Here , a = 3

d = 15 - 3 = 12

putting value of formula

${a}^{54} = a \: + (n - 1)d$

$= > 3 +( 54 - 1) \times 12$

$= > 3 + 53 \times 12$

$= > 3 + 636$

$= > 639$

$so \: {54}^{th} \: team = > {a}^{54} = 639$

we need to calculate team which is 132 more then 54th team

i. e should be 639 + 132 => 771

Therefore an = 771

we need to find n

an = a +( n - 1 ) d

771 = 3 + ( n - 1 ) × 12

771 - 3 = (n - 1 ) × 12

768 = (n - 1 ) × 12

$\frac{768}{12} = n \: - 1$

64 = n - 1

n - 1 = 64

n = 64 + 1

so, 64th team is 132 more than 54th team

Hope it's help you....