Question

1.. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

2.. Which term of the AP:3,15,27,39,
will be 132 more than its 54th term?
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Answer 1

Step-by-step explanation:

1.)

We have,

$\sf{a_{17}-a_{10}=7}$

$\implies{\sf{(a+16d)-(a+9d)=7}}$

$\implies{\sf{a+16d-a-9d=7}}$

$\implies{\sf{16d-9d=7}}$

$\implies{\sf{7d=7}}$

$\implies{\sf{d=\cancel{\frac{7}{7}}}}$

$\implies{\boxed{\sf{d=1}}}$

Therefore, common difference, d = 1

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2.)

We have,

$\boxed{\sf{A.P.\to3,15,27,39...}}$

From the A.P.,

First term, a = 3

Common difference, d = 15 - 3 = 12

Let the nth term be a_n which is 132 more than 54th term(say a_54 ),

Now acc. to the question,

$\sf{a_{n}=a_{54}+132}$

$\implies{\sf{a_{n}=(a+53d)+132}}$

$\implies{\sf{a_{n}=(3+53\times{12})+132}}$

$\implies{\sf{a_{n}=(3+636)+132}}$

$\implies{\sf{a_{n}=639+132}}$

$\implies{\sf{a_{n}=771}}$

Now,

$\sf{a_{n}=a+(n-1)d}$

$\implies{\sf{771=3+(n-1)(12)}}$

$\implies{\sf{771-3=(n-1)(12)}}$

$\implies{\sf{\frac{768}{12}=n-1}}$

$\implies{\sf{64+1=n}}$

$\implies{\boxed{\sf{n=65}}}$

Therefore, the 65th term is 132 greater than 54th term...✓✓✓

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Answer 2

Answer:

Let a be the first term and d be the common difference of A.P.

17th term of A.P t17 = a + 16d.

10th term of A.P is t10 = a + 9d.

Given that

17th term of an A.P exceeds its 10th term by 7.

a + 16d = a + 9d + 7

7d = 7

∴ d = 1