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The sum of the first n terms of an arithmetic progression is given by S_n = n2 + 3n. a) Find the first term. b) Find the common difference. c) Find the 10th term.

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3 parts
a

T1=4T_1 = 4

b

d=2d = 2

c

T10=22T_{10} = 22

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I can see question 7 on the image. Here's the solution:

7) The sum of the first nn terms of an arithmetic progression is given by Sn=n2+3nS_n = n^2 + 3n.

Part a· 1 step

Find the first term.

The first term, T1T_1, is equal to the sum of the first term, S1S_1.

  1. Substitute n=1n=1 into the formula for SnS_n.

    S1=(1)2+3(1)S_1 = (1)^2 + 3(1) S1=1+3S_1 = 1 + 3 S1=4S_1 = 4

    The first term is 4.

    T1=4\boxed{T_1 = 4}

Part b

d=2d = 2

Find the common difference.

To find the common difference, we need the first two terms. We already have T1=4T_1 = 4.

  1. Find the sum of the first two terms, S2S_2.

    S2=(2)2+3(2)S_2 = (2)^2 + 3(2) S2=4+6S_2 = 4 + 6 S2=10S_2 = 10

  2. Find the second term, T2T_2, using S2=T1+T2S_2 = T_1 + T_2.

    T2=S2−S1T_2 = S_2 - S_1 T2=10−4T_2 = 10 - 4 T2=6T_2 = 6

  3. Calculate the common difference, d=T2−T1d = T_2 - T_1.

    d=6−4d = 6 - 4 d=2d = 2

    The common difference is 2.

    d=2\boxed{d = 2}

Part c

T10=22T_{10} = 22

Find the 10th term.

The formula for the nn-th term of an arithmetic progression is Tn=a+(n−1)dT_n = a + (n-1)d, where aa is the first term and dd is the common difference.

  1. Substitute a=4a=4, d=2d=2, and n=10n=10 into the formula.

    T10=4+(10−1)2T_{10} = 4 + (10-1)2 T10=4+(9)2T_{10} = 4 + (9)2 T10=4+18T_{10} = 4 + 18 T10=22T_{10} = 22

    The 10th term is 22.

    T10=22\boxed{T_{10} = 22}
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