Question

Which term of the AP 3 , 15 , 27 , 39 ,.. will be 120 more than its 21st term ?

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Answer 1

✬ Required Term = 31st ✬

Step-by-step explanation:

Given:

• AP series is 3 , 15 , 27 , 39..

To Find:

• Which term will be 120 more than 21st term ?

Solution: Let nth term will be 120 more than 21st term.

Here we have

• a = 3

• d = 15 – 3 = 12

• n = 21

As we know that , an AP series is given by

★ a + (n – 1) × d ★

➙ 3 + (21 – 1) × 12

➙ 3 + 20 × 12

➙ 243

[ Adding 120 to 21st term ]

• 243 + 120 = 363

$\implies{\rm }$ a + (n – 1)d = 363

$\implies{\rm }$ 3 + (n – 1)12 = 363

$\implies{\rm }$ 3 + 12n – 12 = 363

$\implies{\rm }$ 12n = 363 + 9

$\implies{\rm }$ n = 372/12

$\implies{\rm }$ n = 31

Hence, 31st term of AP will be 120 more than 21st term.

Answer 2

Given :-

AP 3 , 15 , 27 , 39

To Find :-

The term will be 120 more than its 21st term

Solution :-

Firstly lets take they are AP or not

15 - 3 = 27 - 15

12 = 12

So,

They are in AP and have common difference 12

Now

$\sf a+(n-1)d$

$\sf 3 + (21-1)12$

$\sf 3+(20)12$

$\sf 3 + (20 \times 12)$

$\sf 3 + 240$

$\sf 243$

Now the required term will be ''x''

$\sf a(x-1)d = 243+120$

$\sf 3+(x-1)12 = 243+120$

$\sf 3 + (x-1)12 = 363$

$\sf (x-1)12=363-3$

$\sf (x-1)12=360$

$\sf x-1 = \dfrac{360}{12}$

$\sf x-1=30$

$\sf x=30+1$

$\sf x=31$