Question

Urgent
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​

Answer 1

Answer:

here is your answer ok thankx

Answer 2

Answer:

1043

Step-by-step explanation:

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)384/3 = 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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