Sum of first n terms is 3n2 + n. Find the 22nd term.
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An AP 5, 8, 11…has 40 terms. Find the last term. Also find the sum of the last 10 terms.
First Term of the AP (a) = 5 Common difference (d) = 8 - 5 = 3 Last term = a40 = a + (40 - 1) d = 5 + 39 × 3...
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In an A.P, cum of four consecutive terms is 28 and their sum of their square is 276. find the four number
Let the numbers be: a - 3d, a - d, a + d, a + 3d a - 3d + a - d + a + d + a + 3d...
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Sum of first n terms is 3n2 + n. Find the 22nd term.
Pls write full question.Your question is a bit ambiguous.
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Sum of first n terms is 3n2 + n. Find the 22nd term.
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Sum of first n terms is 3n^2 + n. Find the 22nd Term.
Sn = 3n2 + n S22 = 3(22)2 + 22 S22 = 1452 + 22 ∴ S22 = 1474 S21 = 3(21)2 + 21 S21 = 1344 a22 = S22 - S21 ∴ a22 = 1474 - 1344 ∴ a22 = 130 Hence, the 22nd term of A.P is 130
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The sum of three numbers is 2. If twice the second number is added to the sum of first and third, we get 1. On adding the sum of second and third numb
Correct Answer - 1,-1,2
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Match the following 1. Summer Solstice (a) 22nd December
1. (c) 2. (e) 3. (a) 4. (b) 5. (d)
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Predict the output. 1. int sum = 10, ctr = 5; sum = sum + ctr–;
1. 15 2. 16
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A company got its Certificate of iIncorporation on 24th August 2010 and on the certificate the date is written as 22nd August 2010. Company allotted s
Yes, it is valid although the registrar issued the certification on 24th but on the certificate 22nd August is written. So in the eyes of law, the company was registered...
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Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution: As the 7th term exceeds the 5th term by 12, so the 5th term will exceed the 3rd term by 12 as well So, n3 = 16 n5 = 28 n7 = 40 n4 or n6 can be...
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