Question

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Answer 1

Answer:

1. 4

2. 28

3.  249

4. ???

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Solutions :-

1) The given sequence = 12,10,8,6,...

First term = 12

Common difference = 10-12 = -2

=> 8-10=-2

=> 6-8=-2

Since the common difference is same throughout the sequence

=> 12,10,8,6,... are in the AP.

We know that

nth term of an AP = an = a+(n-1)d

Now,

24th term = a 24

=> 12+(24-1)(-2)

=> 12+(23)(-2)

=> 12+(-46)

=> 12-46

=> -34

24th term of the given sequence is -34

2)

Given sequence is 8,13,18,23,...

First term = 8

Common difference = 13-8= 5

=> 18-13=5

=> 23-18=5

Since the common difference is same throughout the sequence

=> 8,13,18,23,... are in the AP.

Let nth term = 402

We know that

nth term of an AP = an = a+(n-1)d

Now,

an = 402

=> 8+(n-1)(5) = 402

=> 8+5n-5 = 402

=> 5n+3 = 402

=> 5n = 402-3

=> 5n = 399

=> n = 399/5

n can not be rational number it should be a positive integer.

402 is not a term in the given sequence.

3)

Given AP is 4,9,14,...,254

First term = 4

Common difference =9-4 = 5

Let nth term = 254

We know that

nth term of an AP = an = a+(n-1)d

Now,

an = 254

=> 4+(n-1)(5) = 254

=> 4+5n-5 = 254

=> 5n-1 = 254

=> 5n = 254+1

=> 5n = 255

=> n = 255/5

=>n = 51

254 is the 51st term of the AP

Now 10th term from the end = 51-10+1 = 42

42th term = a+41d

=> 4+(41)(5)

=> 4+205

=> 209

The 10th term form the end of the given AP is 209

4)

Given that an = 3n²-6

Put n = 1 then

a1 = 3(1)²-6

=> a1 = 3(1)-6

=> a1 = 3-6

=> a1 = -3

Put n=2 then

a2 = 3(2)²-6

=> a2 = 3(4)-6

=> a2 = 12-6

=> a2 = 6

Put n = 3 then

a3=3(3)²-6

=> a3 = 3(9)-6

=> a3 = 27-6

a3 = 21

Put n=4 then

a4 = 3(4)²-6

=> a4 = 3(16)-6

=> a4 = 48-6

=> a4 = 42

The four terms of the sequence are -3,6,21,42

5)

Given AP is 16,11,6,1

First term = 16

Common difference =11-16 = -5

We know that

nth term of an AP = an = a+(n-1)d

Now,

12 th term = a12

=> a12 = a+(12-1)d

=> a12 = a+11d

=>a12 = 16+11(-5)

=> a12 = 16+(-55)

=>a12 = 16-55

=> a12 = -39

12th term of the given AP = -39

6)

Given AP is 4,7,10,13,...148

First term = 4

Common difference =7-4 = 3

Let nth term = 148

We know that

nth term of an AP = an = a+(n-1)d

Now,

an = 148

=> 4+(n-1)(3) = 148

=> 4+3n-3 = 148

=> 3n+1 = 148

=>3n = 148-1

=> 3n = 147

=> n = 147/3

=>n = 49

Number of terms in the AP = 49

Used formulae:-

• nth term of an AP = an = a+(n-1)d