Question

If the sum of the first n terms of an AP is given by Sn = (3n² - n), then its 20th term is :
i)152
ii)154
iii)114
iv)116​

Answer 1

Given ,

Sum of nth term of an AP (Sn) = 3(n)² - n

We know that ,

$\boxed{ \sf{a_{n} = S_{n} - S_{n - 1 } }}$

Thus ,

First term = 2 - 0 = 2

Second term = 10 - 2 = 8

$\therefore$ The common difference will be

d = 9 - 2 = 8

Now , the nth term of an AP is given by

$\boxed{ \sf{ a_{n} = a + (n - 1)d}}$

Thus ,

20th term = 2 + (20 - 1) × 8

20th term = 2 + 19 × 8

20th term = 2 + 152

20th term = 154

Therefore , the correct option is (ii) i.e 154

Answer 2

Answer:

The correct answer is option (iv) 116

That is, the 20th term is 116

Step-by-step explanation:

Arithmetic Progression:

An arithmetic sequence or progression is defined as a sequence of numbers in which the difference between two consecutive terms is always a fixed constant. This fixed constant that is added to any term to get the next term in the sequence is known as Common Difference (d). The first term in the sequence is denoted by 'a' and the number of terms is denoted by 'n'. Then, the arithmetic sequence can be written as

a, a+d, a+2d, a+3d, ......., a+(n-1)d

General form or nth term of an Arithmetic sequence:

Tₙ = a + (n - 1) d

Sum of n terms of an Arithmetic Progression:

The sum of the first n terms of an arithmetic progression can be found using the formula,

Sₙ = n/2 [ 2a + (n - 1) d ]

where n ⇒ number of terms

          a ⇒ first term

          d ⇒ common difference

Given:

Sum of first n terms of an AP, $S_{n} = 3n^{2} - n$

Find:

20th term, $T_{20}$ = ?

Solution:

Given that, $S_{n} = 3n^{2} - n$

Sum of first 1 term or $T_{1} = 3 \times 1^{2} - 1$

                                        = $2$

First term, a = 2

Now, Sum of the first 20 terms, $S_{20} = 3\times 20^{2} - 20$

                                                    = $1200-20$

                                                    = $1180$

We know that, Sum of first n terms can be written as,

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

Here,

$S_{20} = \frac{20}{2} [ \ 2a+(20-1)d \ ] = 1180$

⇒ $10 [ \ 2a+19d \ ] = 1180$

⇒ $2a +19d =\frac{1180}{10}$

⇒ $2a+19d = 118$

⇒ $a+a+19d=118$

⇒ $a+T_{20} =118$                 [ ∵ $T_{20} = a+19d$ ]

⇒ $T_{20} = 118 -a$

⇒ $T_{20} = 118 -2$                 [ ∵ $a=2$ ]

⇒ $T_{20} = 116$

That is, 20th term = 116

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