Question

Solve this problem it is worth 50 points, need a good answer, please help me I have my board exam on Thursday

Find the middle term of the sequence formed by all numbers between 9 and 94 which leaves a remainder 1 when divided by 3 .Also find the sum of the numbers on both sides of the middle term respectively

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Answer 1

Answer:

Step-by-step explanation:

9 and 94 .

So 10 , 11 ......... 93 .

remainder 1 when divided by 3 .

Eg : 4 divided by 3 leaves 1 .

10 , 13 , 16 .... 94

d = 3

a = 10

an = ( a + (n-1) d )

= 94 = 10 + 3n - 3

= 3 n = 87

= n = 29

now middle term = 15th

T15= a + (15 - 1)d = 10 + 14 ×3 = 52

Hence, 15th term is 52

So, sum of first 14 term  = 14/2 [ 2 × 10 + (14 -1)×3 ]

= 7[20 + 39] = 7 × 59 = 413

Sum of last 14 term  = S29 -[ S14 + T15 ]

= 29/2[2×10 + (29-1) × 3 ] - 413 - 52

= 29/2[20 + 84 ] - 465

= 1508 - 465

= 1043

Answer 2

   

Good question.

Middle term is 52 and the sum of terms on both sides of the middle term is 104.

Given is the simplest methods and their proof of how these methods are applicable for any APs and how I got them. The proofs are for further reading and not necessary. Just for justification.

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After 9, 10 is the lowest number which leaves remainder 1 on division by 3. So it is the first term.

Also, 94 leaves remainder 1 on division by 3. So it is the last term.

But don't worry about the question. This is just so simple and it doesn't require any much formulas. See below.

We got that 10 is the first term and 94 is the last term.

To find the middle term, add both the first term and last term, and then divide it by 2! That's all!

But why this simple method is applicable to all arithmetic progressions?!

Let the first term be a.

So the last term, i.e., the nth term, becomes a + (n - 1)d.

Okay, take their sum.

$a+a+(n-1)d \\ \\ 2a+(n-1)d$

Okay. Now, take its half.

$\frac{2a+(n-1)d}{2} \\ \\ a+\frac{(n-1)d}{2} \\ \\ a+(\frac{n-1}{2})d$

We got this. This, is  $[\frac{n+1}{2}]^{th}$  term.

How?!

Take $\frac{n+1}{2}$  in the place of n.

So the algebraic expression,

$a+(\frac{n+1}{2}-1)d \\ \\ a+(\frac{n+1-2}{2})d \\ \\ a+(\frac{n-1}{2})d$

Yes! We got the sum!

Hence we proved that half of sum of first term and last term of ANY AP is the middle term.

$[\frac{n+1}{2}]^{th}$  term is the median, i.e., middle term, and also mean.

Okay. Let's find the answer.

Here, the first term and last term are 10 and 94 respectively.

So the middle term is,

$\frac{10+94}{2} = \frac{104}{2} = \bold{52}$

So we get the middle term.

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Next we have to find the sum of both terms on the either sides of the middle term, right?

Nothing. This is equal to the sum of first and last terms!!!

Also twice the middle term!!!

How it could be?!

Okay, let's prove.

Sum of first and last terms = $2a+(n-1)d$

Twice the middle term is also this. We've seen earlier. We've found the middle term by taking half of this.

Middle term is $a+(\frac{n-1}{2})d$.

The left term of this median is the common difference subtracted from median, we know this. So it becomes,

$a+(\frac{n-1}{2})d-d \\ \\ a+(\frac{n-1}{2}-1)d \\ \\ a+(\frac{n-1-2}{2})d \\ \\ a+(\frac{n-3}{2})d$

Also we know that the right term is common difference added to the median. So,

$a+(\frac{n-1}{2})d+d \\ \\ a+(\frac{n-1}{2}+1)d \\ \\ a+(\frac{n-1+2}{2})d \\ \\ a+(\frac{n+1}{2})d$

Adding the both, we get,

$a+(\frac{n-3}{2})d+a+(\frac{n+1}{2})d \\ \\ 2a+(\frac{n-3}{2}+\frac{n+1}{2})d \\ \\ 2a+(\frac{n-3+n+1}{2})d \\ \\ 2a+(\frac{2n-2}{2})d \\ \\ 2a+(n-1)d \\ \\ a+a+(n-1)d \\ \\ 2[a+(\frac{n-1}{2})d]$

Hence proved!!!

So, the sum of terms on both sides of the middle term is 10 + 94 = 2 × 52 = 104.

This is only for further reading. I've posted this just to know everyone. Also, this question is of higher points. So how can I answer it in one or two words???!!!

Sorry for my huge answer if it distracted you.

Hope my answer will be helpful to you.

Don't forget to mark it as the brainliest if this answer helps you.

Also, please ask me if you have any doubts.

Thank you. Have a nice day. :-))

#adithyasajeevan