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find the middle term of the sequence formed by all three digits which leave a remainder 3 when divided by 4 also find the sum of all numbers on both side of the middle term separately

Posted by Vbdv Dgff 5 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 5 years, 10 months ago

The three digit numbers which leaves 3 as remainder when divided by 4 are : 103,107, 111,..... 999
Now, the first number $a = 103$ , last number l = 999
and common difference d = 4
Let the number of terms in this sequence be $n$ .
$l = a + (n - 1)d$
or, $103 + ( n - 1)4 = 999$
(n -1)4 = 999 - 103
$(n -1)4 = 896$
$(n - 1)$ = $\frac{896}{4}$ = 224
$n = 224 + 1 = 225$
Middle term = $\frac { 225 + 1 } { 2 }$
= 113^th term
a₁₁₃ = 103 + 112 $\times$ 4 = 551
and a₁₁₂ = 551 - 4 = 547
Sum of first 112 terms
$= \frac { 112 } { 2 } ( 103 \times 2 + 111 \times 4 )$
$= \frac { 112 } { 2 } ( 206 + 444 )$
$= 56 \times 650$
= $36400$
and a₁₀₄ = $551 + 4 = 555$
The Sum of last 112 terms
= $\frac { 112 } { 2 } ( 2 \times 555 + 111 \times 4 )$
$= 56 ( 1110 \times 444 )$
$= 56 \times 1554$
= 87024

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