Question

The sum of first ‘n’ terms of an AP is given by Sn = 3n square - n. Find AP.
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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{A.P=2,8,14,20,.....}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies s_{n} = {3n}^{2} - n \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies A.P= ?$

• According to given question :

$\tt \circ \: s_{n} = {3n}^{2} - n \\ \\ \tt\because \: N= 1,2,3,.... \\ \\ \bold{As \: we \: know \: that(n = 1)} \\ \tt:\implies s_{1} = 3 \times {1}^{2} -1 \\ \\ \tt:\implies s_{1} =3 - 1 \\ \\ \green{\tt:\implies s_{1} =2 = a_{1} } \\ \\ \bold{For \: (n = 2)} \\ \tt:\implies s_{2} =3 \times {2}^{2} - 2 \\ \\ \tt:\implies s_{2} =3 \times 4 - 2 \\ \\ \tt:\implies s_{2} =12 - 2 \\ \\ \green{\tt:\implies s_{2} =10} \\ \\ \bold{For \: second \: term : } \\ \tt:\implies a_{2} = s_{2} - s_{1} \\ \\ \tt:\implies a_{2} =10 - 2 \\ \\ \green{\tt:\implies a_{2} =8} \\ \\ \bold{For \:( n = 3)} \\ \\ \tt:\implies s_{3} =3 \times {3}^{2} - 3 \\ \\ \tt:\implies s_{3} =3 \times 9 - 3 \\ \\ \tt:\implies s_{3} =27 - 3 \\ \\ \green{\tt:\implies s_{3} =24} \\ \\ \bold{For \:third \: term : } \\ \tt:\implies a_{3} = s_{3} - s_{2} \\ \\ \tt:\implies a_{3} = 24 - 10 \\ \\ \green{\tt:\implies a_{3} =14} \\ \\ \green{\tt \therefore A.P = 2,8,14,20,......}$

Answer 2

$\bigstar\ \Large\underline{\bold{\tt{GIVEN :}}}$

The sum of the first n terms of an A.P. is given by :-

$\displaystyle{\tt{S_n=3n^2-n}}$

$\bigstar\ \Large\underline{\bold{\tt{TO\ FIND :}}}$

The A.P.

$\bigstar\ \Large\underline{\bold{\tt{ACKNOWLEDGEMENT :}}}$

$\displaystyle{\tt{S_n-S_{n-1}=a_n}}$

$\displaystyle{\tt{\implies S_{n-1}=3(n-1)^2-(n-1)}}$

$\bigstar\ \Large\underline{\bold{\tt{SOLUTION :}}}$

For a₁,

a₁ = S₁ - S₍₁₋₁₎

⇒a₁ = 3(1)² - (1) - [3(1 - 1)² - (1 - 1)]

⇒a₁ = 3 - 1 - (0 - 0)

⇒a₁ = 3 - 1

⇒a₁ = 2

$\rule{176}{2}$

For a₂,

a₂ = S₂ - S₍₂₋₁₎

⇒a₂ = 3(2)² - 2 - [3(2 - 1)² - (2 - 1)]

⇒a₂ = 12 - 2 - (3 - 1)

⇒a₂ = 10 - 2

⇒a₂ = 8

$\rule{176}{2}$

For a₃,

a₃ = S₃ - S₍₃₋₁₎

⇒a₃ = 3(3)² - 3 - [3(3 - 1)² - (3 - 1)]

⇒a₃ = 27 - 3 - (12 - 2)

⇒a₃ = 24 - 10

⇒a₃ = 14

$\rule{176}{2}$

Now, we get d = a₂ - a₁ = a₃ = a₂ = 6

So, the AP is 2, 8, 14, 20, 26, ......