Question

Let s₁, s₂, s₃,…..,s₁₀ respectively be the sum to 12 terms of 10 A.P.s whose first terms are 1,2,3, …. ,10 and the common differences are 1, 3, 5, …. , 19 respectively. Then  \(10∑^{10}_{i=1} s_i\) is

• A. 7260
• B. 7220
• C. 7360
• D. 7380

Answer 1

The Correct Option is A

We are given 10 arithmetic progressions (A.P.s), each with 12 terms. The first term of each A.P. is \(a_i\), where \(a_1 = 1, a_2 = 2, ..., a_{10} = 10\). The common differences \(d_i\) for these A.P.s are given as 1, 3, 5, ..., 19 respectively. We need to find the value of \(10∑^{10}_{i=1} s_i\), where \(s_i\) is the sum of the first 12 terms of i-th A.P.

The sum \(s_i\) of the first 12 terms of an A.P. is given by the formula:

s_i = \frac{n}{2} [2a_i + (n-1)d_i]

Here, \(n = 12\). Substitute this into the formula:

s_i = \frac{12}{2} [2a_i + 11d_i] = 6 [2a_i + 11d_i]

Now we calculate each \(s_i\) for \(i = 1\) to 10:

• \(s_1 = 6[2 \times 1 + 11 \times 1] = 6[2 + 11] = 6 \times 13 = 78\)
• \(s_2 = 6[2 \times 2 + 11 \times 3] = 6[4 + 33] = 6 \times 37 = 222\)
• \(s_3 = 6[2 \times 3 + 11 \times 5] = 6[6 + 55] = 6 \times 61 = 366\)
• \(s_4 = 6[2 \times 4 + 11 \times 7] = 6[8 + 77] = 6 \times 85 = 510\)
• \(s_5 = 6[2 \times 5 + 11 \times 9] = 6[10 + 99] = 6 \times 109 = 654\)
• \(s_6 = 6[2 \times 6 + 11 \times 11] = 6[12 + 121] = 6 \times 133 = 798\)
• \(s_7 = 6[2 \times 7 + 11 \times 13] = 6[14 + 143] = 6 \times 157 = 942\)
• \(s_8 = 6[2 \times 8 + 11 \times 15] = 6[16 + 165] = 6 \times 181 = 1086\)
• \(s_9 = 6[2 \times 9 + 11 \times 17] = 6[18 + 187] = 6 \times 205 = 1230\)
• \(s_{10} = 6[2 \times 10 + 11 \times 19] = 6[20 + 209] = 6 \times 229 = 1374\)

Now find the sum:

\sum_{i=1}^{10} s_i = 78 + 222 + 366 + 510 + 654 + 798 + 942 + 1086 + 1230 + 1374

Calculate the total sum:

78 + 222 + 366 + 510 + 654 + 798 + 942 + 1086 + 1230 + 1374 = 726

Finally, we multiply this result by 10:

10 \times 726 = 7260

Thus, the value of 10\sum^{10}_{i=1} s_i is 7260.