Math, asked by anantrajusharma, 7 months ago

Find the sum of all two-digit numbers which when divided by 4, yields 1 as remainder.

Answers

Answered by AnantSharmaGUNA
1

Answer:

1210

Step-by-step explanation:

The two - digit numbers which when divided by 4, yield 1 as remainder, are as follows: -

13, 17, … 97

Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with first term 13 and common difference 4.

Last Term of the A.P(l) = 97

Let n be the number of terms of the A.P.

It is known that the nth term of an A.P. is given by -

an = a + (n –1) d

⇒ 97 = 13 + (n –1) (4)

⇒ 4 (n –1) = 84

⇒ n – 1 = 21

⇒ n = 22

Sum of n terms of an A.P. is given by -

Sn = (n/2)[a + l]

∴ S22 = (22/2)[13 + 97]

= 11 × 110

= 1210

Thus, the required sum is 1210.

JAI SHREE KRISHNA

Answered by legend123890
0

Answer:

Solution : The two-digit numbers, which when divided by 4, yield 1 as remainder, are, 13,17,…97. This series forms an A.P. with first term, a=13 and common difference, d=4.

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