Question

Find the sum of all the non-
negative terms of the following
sequence 100, 97, 94,...... ​

Answer 1

Answer:

All are in A.P

we know this as

common difference is same !

Given:

a = t1 = 100

t2 = 97

d = 97-100 = -3

tn = 1

it is zero because question want sum of non - negative terms

so it will not go beyond zero

tn = a +(n-1)d

1= 100 + -3n +3

-102 = -3n

n = 34

Sn = n/2 [ 2a +(n-1) d]

= 34/2 [200 + 33 × (-3) ]

= 34/2 [ 200-99]

= 34 × 51 ..........(50.5~ 51)

= 1734

Hope it helps yoy!

Answer 2

Answer:

Total Sum = 1717

Step-by-step explanation:

Difference(d) = -3

1st Term(a) = 100

First Let's calculate total terms:

We can assume that Last term will be 1

Last Term (l) = 1

Using formula AP = a + (n-1)d

1 = 100 + (n-1)-3

1 = 100 - 3n + 3

1 = 103 - 3n

1 - 103 = -3n

-102 = -3n

n =  102/3

n = 34...total terms

So Now we have total terms...n = 34;

Let's now use the formula to calculate total sum...

Sum = n/2 (a + l)

So Putting the Values in the formula:

S = 34/2(100 + 1)

S = 17(100 + 1)

S = 1700 + 17

S = 1717...Ans

Thankyou I hope now the question is clear...

In case you are having doubt how did I assumed '1' as last term

you can see yourself

100 = 1st Term

100 - 3 = 97...2nd Term

97 - 3 =  94 = 3rd term

94-3 = 91 = 4th term

...

If you see in this there are 4 terms in every 10 digits and as 100 to 90 one is grouped, every other would be grouped like this and we only need positive...So We took 1 to 10, and as 91  was the last in 100 to 90 , 1 would be the last term... in 1 to 10 hence the 1 is last term

Thanks... :D