Question

zero?
10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference
11. Which term of the AP:3, 15, 27, 39,... will be 132 more than its 54th term?​

Answer 1

Answer:

10. 1

11. 65

Step-by-step explanation:

10.

-----------------------------

• a₁₇-a₇=7
• a₁₇=a₁+16d
• a₁₀=a₁+9d
• a₁₇-a₇= a₁+16d - (a₁+9d)= 7d
• 7d=7 => d=1

11.

-----------------------------

AP: 3, 15, 27, 39

a₁=3

d=15-3=12

a₅₄= 3+53d= 3+53*12= 639

aₙ=3+(n-1)d= 3+12n-12= 12n-9

• a₅₄+132= 639+132= 771

• 12n-9=771
• 12n=780
• n=780/12
• n=65

Answer 2

Answer:

10. Common difference = 1

11. 65 = n

Step-by-step explanation:

10. As we know,

a17 = a+(17-1)d

a17 = a+16d

and,

a10 = a+(10-1)d

a10 = a+9d

A.T.Q

a17-a10 = 7

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

a+16d-a-9d = 7

16d-9d = 7

7d = 7

d = 7/7

d = 1.

11. Given,

a = 3

d = 15-3 = 12

As we know,

a54 = a+(54-1)d

a54 = 3+53*12

a54 = 3+636

a54=639

Now 639+132=771

an = 3+(n-1)12

771-3 = (n-1)12

768 = (n-1)12

768/12 = n-1

64+1 = n

65 =n

Hence, the 65th term of AP will be 132 more than 54th term..

Hope it helps....