Question

if the sum of first n terms of an AP is 3n2 - 2n, find the AP and its 19 term?​

Answer 1

Answer:

 $\bigstar{\bold{The\:A.P\:is1,7,13....}}$

 $\bigstar{\bold{The\:19th\:term\:(a_{19})\:of\:the\:A.P\:is\:109}}$

Explanation:

 $\Large{\underline{\underline{\bf{Given:}}}}$

• Sum of first n terms of the A.P = 3n² - 2n

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The A.P

• The 19th term of the A.P

$\Large{\underline{\underline{\bf{Solution:}}}}$

➜ First we have to find the A.P

➜ Here sum of n terms is given as 3n² - 2n

➜ Hence sum of first term = First term (a₁)

➜ First term = 3 × 1² - 2 × 1

   First term = 3 - 2

   First term = 1

➜ Hence first term of the A.P is 1

➜ Now the sum of 2 terms is given by

    Sum of two terms = 3 × 2² - 2 × 2

    Sum of two terms = 3 × 4 - 4

    Sum of two terms = 12 - 4

    Sum of two terms = 8

➜ Now the second term (a₂) is given by,

    Second term = Sum of two terms - First term

➜ Substitute the data,

    Second term = 8 - 1  

    Second term = 7

➜ Now we have to find the common difference (d) of the A.P

   Common difference (d) = a₂ - a₁

➜ Substitute the data,

    d = 7 - 1

    d = 6

➜ Hence common difference of the A.P is 6

➜ Now third term of the A.P is given by,

    a₃ = a₂ + d

     a₃ = 7 + 6

     a₃ = 13

➜ Hence the A.P is 1, 7, 13.....

     $\boxed{\bold{The\:A.P\:is1,7,13....}}$

➜ Now we have to find the 19th term (a₁₉) of the A.P

➜ The 19th term of the A.P is given by,

   a₁₉ = a₁ + 18d

➜ Substitute the data,

    a₁₉ = 1 + 18 × 6

   a₁₉ = 1 + 108

   a₁₉ = 109

➜ Hence the 19th term of the A.P is 109.

     $\boxed{\bold{The\:19th\:term\:(a_{19})\:of\:the\:A.P\:is\:109}}$

 $\Large{\underline{\underline{\bf{Notes:}}}}$

➠ The sum of n terms of an A.P is given by,

     $\sf{S_n=\dfrac{n}{2} (a_1+a_n)}$

    $\sf{S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)}$

 

Answer 2

Answer:

\bigstar{\bold{The\:A.P\:is1,7,13....}}★TheA.Pis1,7,13....

\bigstar{\bold{The\:19th\:term\:(a_{19})\:of\:the\:A.P\:is\:109}}★The19thterm(a

19

)oftheA.Pis109

Explanation:

\Large{\underline{\underline{\bf{Given:}}}}

Given:

Sum of first n terms of the A.P = 3n² - 2n

\Large{\underline{\underline{\bf{To\:Find:}}}}

ToFind:

The A.P

The 19th term of the A.P

\Large{\underline{\underline{\bf{Solution:}}}}

Solution:

➜ First we have to find the A.P

➜ Here sum of n terms is given as 3n² - 2n

➜ Hence sum of first term = First term (a₁)

➜ First term = 3 × 1² - 2 × 1

First term = 3 - 2

First term = 1

➜ Hence first term of the A.P is 1

➜ Now the sum of 2 terms is given by

Sum of two terms = 3 × 2² - 2 × 2

Sum of two terms = 3 × 4 - 4

Sum of two terms = 12 - 4

Sum of two terms = 8

➜ Now the second term (a₂) is given by,

Second term = Sum of two terms - First term

➜ Substitute the data,

Second term = 8 - 1

Second term = 7

➜ Now we have to find the common difference (d) of the A.P

Common difference (d) = a₂ - a₁

➜ Substitute the data,

d = 7 - 1

d = 6

➜ Hence common difference of the A.P is 6

➜ Now third term of the A.P is given by,

a₃ = a₂ + d

a₃ = 7 + 6

a₃ = 13

➜ Hence the A.P is 1, 7, 13.....

\boxed{\bold{The\:A.P\:is1,7,13....}}

TheA.Pis1,7,13....

➜ Now we have to find the 19th term (a₁₉) of the A.P

➜ The 19th term of the A.P is given by,

a₁₉ = a₁ + 18d

➜ Substitute the data,

a₁₉ = 1 + 18 × 6

a₁₉ = 1 + 108

a₁₉ = 109

➜ Hence the 19th term of the A.P is 109.

\boxed{\bold{The\:19th\:term\:(a_{19})\:of\:the\:A.P\:is\:109}}

The19thterm(a

19

)oftheA.Pis109

\Large{\underline{\underline{\bf{Notes:}}}}

Notes:

➠ The sum of n terms of an A.P is given by,

\sf{S_n=\dfrac{n}{2} (a_1+a_n)}S

n

=

2

n

(a

1

+a

n

)

\sf{S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)}S

n

=

2

n

(2a

1

+(n−1)×d)