Question

The sum of all two digit numbers which when divided by 4 yield unity as remainder is​

Answer 1

Answer: Or This series forms an A.P. with first term 13 and common difference 4. Let n be the number of terms of the A.P. Thus, the required sum is 1210.

Step-by-step explanation:

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 13 and common difference 4.

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,

S subscript n space equals space n over 2 space open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets

therefore space S subscript 22 space equals space 22 over 2 space open square brackets 22 open parentheses 13 close parentheses space plus space open parentheses 22 minus 1 close parentheses open parentheses 4 close parentheses close square brackets

space space space space space space space space space space space space space equals space 11 open square brackets 26 space plus space 84 close square brackets

space space space space space space space space space space space space space equals space 1210

Thus, the required sum is 1210.

Answer 2

Solution -:

the number are 13 , 17.........97

this is an arithmetic progression with the first term 13 and common difference for let the number of term Be n .

Then ,,

$\displaystyle{}97 = 13 + (n - 1)4$

$\displaystyle{} \bold{ \because a _n = a + (n - 1)}$

$\displaystyle{} \implies \: 4n = 88 \implies \: n = 22$

therefore the sum of the number

$\displaystyle \: s_n \: = \frac{n}{2}(a + l) </strong><strong> =</strong><strong>\frac{22}{2} 13 + 97$

$\displaystyle \bold{11(110) = 1210}$