Question

given the arithmetic sequence 3,6,9,12 how many terms are there in the sequence if the last term is 153

Answer 1

Step-by-step explanation:

tn=153,a=t1=3,t2=6,t3=9 and t4=12 so, d=3

tn=a+(n-1) d

153=3+(n-1)3

153=3+3n-3

153=3n

n=153/3

n=51

so, there are 51 terms.

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Answer 2

Step-by-step explanation:

Given: 3,6,9,12...153

To find: How many terms are there in A.P.

Solution:

We know that $n^{th}$ term of AP is given by

$\bold{a_n=a+(n-1)d}$

here, a is first term and d is common difference.

In the given AP

a=3

d=3

put these values in the formula

$153 = 3 + (n - 1)3$

$153 - 3 = 3(n - 1)$

$150 = 3(n - 1)$

$\frac{150}{3} = n - 1$

$50 = n - 1$

$n = 50 + 1$

$\bold{n = 51 }$

Final answer:

Number of terms in the AP are 51.

Hope it helps you.

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