Question

which term of AP8,14,20,26...will be 72 more than its 41st term

Answer 1

Answer:

thus 53 rd term is 72 more than its 41st term

Step-by-step explanation:

8,14,20,26....

a=8, d=14-8=6

41st term =a+40d

=8+40*6=240+8=248

adding 72+248=320

let this is nth term

then 8+(n-1)*6=320

(n-1)6=312

n-1=312/6=52

so n=53

thus 53 rd term is 72 more than its 41st term

Answer 2

Answer:

53rd term of given AP will be 72 more than 41st term

Step-by-step explanation:

Tn = a + ( n - 1 ) d

where a = first term = 8

d = common difference = 14 - 8 = 6

n = term number

41st term

Tn = a + ( n - 1 ) d

=> T41 = 8 + ( 41 - 1 ) 6

=> T41 = 8 + (40) 6

=> T41 = 8 + 240

=> T41 = 248

Now add 72

=> 248 + 72

=> 320

Which term of AP is 320 [Tn = 320]

Tn = a + ( n - 1 ) d

=> 320 = 8 + ( n - 1 ) 6

=> 320 = 8 + 6n - 6

=> 320 = 2 + 6n

=> 320 - 2 = 6n

=> 318 = 6n

=> n = 318/6

=> n = 53