Question

which term of the AP 8, 14, 20
,26 will be 72 more than its 41st time

Answer 1

Question:

Which term of the AP 8, 14, 20, 26 will be 72 more than its 41st term?

Solution:

a = 8

d = a2 - a1 = 14 - 8 = 6

Find the nth term:

an = a1 + (n - 1)d

an = 8 + (n - 1)6

an = 8 + 6n - 6

an = 2 + 6n

Find the 41st term:

a41 = 2 + 6(41)

a41 = 248

Find 72 more than 41st term:

Number = 248 + 72 = 320

Find the term that gives 320:

an = 2 + 6n

320 = 2 + 6n

6n = 318

n = 318 ÷ 6 = 53

Answer: The 53rd term will be 72 more than 41st term.

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ALTERNATIVE METHOD:

Find the common difference:

d = a2 - a1

d = 14 - 8

d = 6

Define n:

Let n be the term that gives the sum of 72 more than 41st term

Find the number of terms between n and 41st term:

Number of terms = 72 ÷ 6 = 12

Find n:

n = 41 + 12 = 53

Answer: The 53rd term will be 72 more than 41st term.

Answer 2

hey!

here is answer



given

a1=8

a2=14


a3=20

d=a2-a1.

14-8=6

an= a+(n-1)d

an=8+(n-1)6

an=8+6n-6

an=2+6n

hence


a41=2+6(41)

a41=248

number = 248+72

320

hence

term=


an=2+6n

320=2+6n

6n=320-2

6n=318

n=318/6.

n=53rd term

hope it helps

thanks