Question

please help me... its urgent

The 17th term of an AP exceeds its 10th term by 7. Find the common dife
Which term of the AP:3, 15, 27, 39... will be 132 inore than its 54th term
Two APs have the same common difference.​

Answer 1

a17=a+16d

a10=a+9d

so

(a+16d)-(a+9d)=7

a+16d-a-9d=7

7d=7

d=1

2. find term

an-a54=132

(3+(n-1)12)-(3+(54-1)12)=132

3+12n-12-3-53×12=132

12n= 12+636+132

n= 780/12

n=65

I hope it help u

but why u mix two questions

thank

Answer 2

Que:-

(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 inore than its 54th term.

Solution 1:-

A.T.Q.

a17 = a10 + 7

=) a + 16d = a + 9d + 7

=) 16d - 9d = 7

=) 7d = 7

=) d = 1

Solution 2:-

First, Let's Find the 54th Term of the A.P.

=) a54 = a + 53(12)

=) a54 = 3 + 636

=) a54 = 639

Now,

It is asked that ,Which term of the AP:3, 15, 27, 39... will be 132 inore than its 54th term?

=) 639 + 132

=) 771

Now,

an = a + ( n - 1)d

=) 771 = 3 + ( n - 1)(12)

=) 768/12 = n - 1

=) n = 64 + 1

=) n = 65

Hence,

65th term of the given A.P. will be 132 more than its 54th term.