Math, asked by pandiyakyatish, 16 days ago

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Answered by Anonymous
1

Answer:

53rd term is the answer.

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Answered by tennetiraj86
4

Given :-

The AP is 8,14,20,26,...

To find :-

The term of the AP will be 72 more than ist 41st term.

Solution :-

Given Arithmetic Progression is 8,14,20,26,...

First term (a) = 8

Common difference (d) = 14-8 = 6

We know that

The nth term of an AP is an = a+(n-1)d

41st term = 8+(41-1)(6)

=> 41st term = 8+(40×6)

=> 41st term = 8+240

=> 41st term = 248

Let the term which 72 more than 41st term of the AP be an

Therefore an = 41st term + 72

=> a+(n-1)d = 248+72

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

=> 8+6n-6 = 320

=> 6n+2 = 320

=> 6n = 320-2

=> 6n = 318

=> n = 318/6

=> n = 53

The required term = 53rd term

Answer :-

53rd term of the AP is 72 more than its 41st term.

Check :-

We have,

a = 8

d = 6

41st term = 248

Now,

53rd term = 8+(53-1)(6)

=> 53rd term = 8+(52×6)

=> 53rd term = 8+312

=> 53rd term = 320

=> 53rd term = 248+72

=> 53rd term = 41st term +72

Verified the given relations in the given problem.

Used formulae:-

The nth term of an AP is an = a+(n-1)d

  • an = nth term
  • a = first term
  • d = common difference
  • n = no. of terms
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