Question

that are (i) divisible by 9 ( not div
13. Find the middle
term of the sequence formed by all the
numbers between 9 and 95 which leave a remainder 1
when divided by 3. Also find the sum of numbers on
both the sides of the middle term separately.

Please send me ans as soon as possible its urgent​

Answer 1

The sequence are formed by all the numbers between 9 and 95 , which leaves remainder 1 when it is divided by 3 : 10, 13, 16, .........94
Here you can see 10, 13 , 16 , .....94 are in AP where first term , a = 10 and common difference , d = 94 .
From AP nth term formula ,
Tn = a + (n -1)d
94 = 10 + (n -1)3
84/3 = n -1 ⇒n = 29

Hence, there are 29 terms between 10 and 94
It means middle term is (n + 1)/2 th term
e.g., middle term = (29+1)/2 = 15th term
So, T₁₅ = a + (15 - 1)d = 10 + 14 ×3 = 52
Hence, 15th term is 52

here middle is 15th , it means 14 terms are in left side of it and 14 terms are in right side of it.
So, sum of first 14 term {left side } = 14/2 [ 2 × 10 + (14 -1)×3 ] [∵Sn = n/2[2a + (n-1)d]
= 7[20 + 39] = 7 × 59 = 413

Now, sum of last 14 term { right side of middle term } = S₂₉ -[ S₁₄ + T₁₅ ]
= 29/2[2 × 10 + (29-1) × 3 ] - 413 - 52
= 29/2[20 + 84 ] - 465
= 1508 - 465
= 1043

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