Question

Observe the following pattern for obtaining the sums then find the sum of numbers from 781 to 790?
\[1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55\]
\[11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 155\]
\[21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 = 255\]
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\[51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 = 555\]
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\[101 + 102 + 103 + 104 + 105 + 106 + 107 + 108 + 109 + 110 = 1055\]
\[251 + 252 + 253 + 254 + 255 + 256 + 257 + 258 + 259 + 260 = 2555\]

A. 7055
B. 7855
C. 7955
D. 8155

Answer 1

Hint:Here we observe the given sum of different patterns and try to relate the terms of the series to the sum of the series. Form a general pattern which tells us the answer based on the numbers at ones, tens and hundreds place and then substitute the values from the given number in one, tens and hundreds place.

Complete step-by-step answer:
We look at each series separately.
From the first series \[1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55\]
Here the terms from 1 to 9 have no digit in tens place and the sum is 55.
From the second series \[11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 155\]
Here the terms from 11 to 19 have 1 in the tens place and the sum on RHS has 1 at hundreds place.
From the third series \[21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 = 255\]
Here the terms from 21 to 29 have 2 in the tens place and the sum on RHS has 2 at hundreds place.
From the series \[51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 = 555\]
Here the terms from 51 to 59 have 5 in the tens place and the sum on RHS has 5 at hundreds place.
From the series \[101 + 102 + 103 + 104 + 105 + 106 + 107 + 108 + 109 + 110 = 1055\]
Here the terms from 101 to 109 have 1 in hundreds place and 0 in tens place and the sum on RHS has 1 in thousands place and 0 at hundreds place and 55 is fixed after that.
From the series \[251 + 252 + 253 + 254 + 255 + 256 + 257 + 258 + 259 + 260 = 2555\]
Here the terms from 251 to 259 have 2 in hundreds place and 5 in tens place and the sum on RHS has 2 in thousands place and 5 at hundreds place and 55 is fixed after that.
So, the general from of sum is:
The number in series at hundreds places is the number at thousand places in the sum.
The number in series at the tens place is the number at the hundreds place in the sum.
And the numbers in the sum at ones and tens place will be 55.
So, the sum of the series \[781 + 782 + 783 + 784 + 785 + 786 + 787 + 788 + 789 + 790\]
Here the terms in the series have 7 in hundreds place and 8 in tens place, so the sum in RHS will have 7 in thousands place and 8 in hundreds place and 55 after that.
\[781 + 782 + 783 + 784 + 785 + 786 + 787 + 788 + 789 + 790 = 7855\]

So, the correct answer is “Option B”.

Note:Students are likely to make mistakes if they cannot make up a pattern between the terms in the series and the sum. Many students might try actually to calculate the sum of numbers from 781 to 790 which is a long calculative process and if we are doing that then there is no use of the given sum of series.We can also solve this question by using Arithmetic progression.Using sum of AP series formula i.e $S_n=\dfrac{n}{2}(2a+(n-1)d)$, where $a$=first term=$781$, $d$=common difference=$1$ and $n$= number of terms from the series \[781 + 782 + 783 + 784 + 785 + 786 + 787 + 788 + 789 + 790 = 7855\].By using this formula also we get the same answer.