Question

Help me sos?

Answer 1

D `100162`

Explanation:

Given:

`46+49+52+...+775`

Note that the common difference is `3`, so the number of terms is:

`(775-46)/3 + 1 = 729/3 + 1 = 243+1 = 244`

Since this is an arithmetic series, the average term is the same as the average of the first and last terms, namely:

`(46+775)/2 = 821/2`

So the sum of the series is:

`244 * 821/2 = 122 * 821 = 100162`

Answer 2

See the answer below...

Explanation:

For any AP Series, if `"first term"` `->` a
`" common difference "->d`,

• The `color(red)(n_(th)` term of the series is `color(red)(n_(th) "term"=a+(n-1)d`
• Sum of the series upto `color(red)(n_(th)` term is `color(red)("sum"=n/2{2a+(n-1)d}`

In the given series, `color(red)(a=46` and `color(red)(d=(49-46)=3`
Hence, `" "color(red)(46+(n-1)3=775`
`" "=>3n-3=729=>n=244`

Now, we have the value of `a,d,n` ,so we can easily determine the result of `color(red)("sum"`.

`color(red)("sum")=244/2{2xx46+(244-1)3}=color(green)(100162`

Hence,the correct option id (D)`100162`

Hope it helps...
Thank you...

Answer 3

`D`

Explanation:

`"the sum to n terms of an arithmetic series is"`

`•color(white)(x)S_n=n/2[2a+(n-1)d]`

#"where a is the first term and d "color(blue)"the common difference"#

`"here "a=46" and "d=52-49=49-46=3`

`"the nth term of an arithmetic series is"`

`•color(white)(x)a_n=a+(n-1)d`

`rArr46+3(n-1)=775larrcolor(blue)"solve for n"`

`rArr46+3n-3=775`

`rArr3n=732rArra=732/3=244`

`"we require the sum to 244 terms"`

`rArrS_(244)=122[(2xx46)+(243xx3)]`

`color(white)(rArrS_(244))=100162`

Answer 4

Figure out which term `775` is, add the reverse of the sum to itself, transform it into multiplication, and obtain D. `100162`

Explanation:

Let's set that sum into a variable, say, `M`:

`M = 46 + 49 + 52 + ... + 775`

OK, now let's use the formula for arithmetic sequences to find out which term `775` is, or, well, how many terms are in this sum?

`a_n = a_1 + d(n - 1)`

We know `a_n = 775` since that's the number whose `n` we're currently looking for... we know the first term, `a_1 = 46`... then there's the difference between two terms, say between `a_2` and `a_1`: `d = a_2 - a_1 = 49 - 46 = 3`. Substitute:

`775 = 46 + 3(n - 1)`

Expand `3(n - 1)`:

`775 = 46 + 3n - 3`

Add the constants:

`775 = 43 + 3n`

Subtract both sides by `43`:

`775 - 43 = 43 + 3n - 43`

`732 = 3n`

Divide both sides by `3`:

`732/3 = (3n)/3`

`244 = n`

So now we know that `775` is the `244`th term of the arithmetic sequence. Bigger than expected! Now back to `M`:

`M = 46 + 49 + 52 + ... + 775`

Let's reverse the sum:

`M = 775 + ... + 52 + 49 + 46`

Add those two together:

`M + M = (46 + 775) + ... + (775 + 46)`

The left side simplifies to `2M`, the right side... well, let's reveal a bit more about the sum (by remembering about the common difference, `d = 3`):

`2M = (46 + 775) + (49 + 772) + ... + (772 + 49) + (775 + 46)`

Seems symmetrical! Not only that, but all these terms are equal; as `46` is added by `3` to become `49`, `775` is also decreased by `3` to become `772`. This keeps going until it becomes `(775 + 46)` at the end.

And how many terms are there? `244`.

`2M = (46 + 775)(244)`

See what I did there? Since they're all equal, I could just use one of them, and multiply it over by the amount of terms. However, we are currently solving for `2M`, not `M`, so let's divide by `2` first:

`(2M)/2 = ((46 + 775)(244))/2`

`M = (46 + 775)(122)`

Solve!

`M = (821)(122)`

Ouch. In this case I would usually "split" one of the numbers, usually the smaller one, then expand, as such:

`M = (821)(100 + 20 + 2)`

`M = (821)(100) + (821)(20) + (821)(2)`

`M = 82100 + 16420 + 1642`

`M = 98520 + 1642`

`M = 100162`

Done!