Question

find the sum of first 16 terms of sequence whose nth term is given by Tn= 5n-3​

Answer 1

Step-by-step explanation:

Nth term

Tn = 5n-3

T1 = 5*1-3 = 2

T2 = 5*2-3 = 7

T16 = 5*16-3 = 77

Common difference, d = 7-2=5

$S_{n} = \frac{n}{2} (2a + (n - 1)d) \\ S_{16} = \frac{16}{2} (2a + (16 - 1)d) \\ S_{16} = \frac{16}{2} (2a + (16 - 1)d) \\S_{16} =8(2 \times 2 + 15 \times 5) \\ S_{16} =8(4 + 75) \\ S_{16} =8(79) \\ S_{16} =632$

Answer 2

Concept:

$S_{n} = \frac{n}{2} (2a+(n-1)d)$

Given:

$T_{n} = 5n-3$

Find:

find the sum of first 16 terms

Answer:

$S_{16} = 632$

Step-by-step explanation:

$T_{n} = 5n-3 = A_{n}$

$T_{1} =A_{1}= 5-3=2$

$T_{2} =A_{2}= 10-3=7$

$T_{3}= A_{3}= 15-3=12$

$d= T_{2}- T_{1} =7-2=5\\d= T_{3}- T_{2} = 12-7=5$

$here \\n=16\\d=5\\a=2$

$S_{n}= \frac{n}{2} (2a+(n-1)d)$

$S_{n} = \frac{n}{2}$ $(2a+nd-d)$

$2a=2$×$2=4$

$nd=16$×$5=80$

$S_{16} = \frac{16}{2}$$(4+80-5)$

$S_{16} = 8$×$79$

$S_{16} = 632$

Hence the sum of the first 16 terms is 632

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