Question

Please answer my question as soon as possible please it's urgent I will rate your answer and Mark as brainlist ​

Answer 1

Step-by-step explanation:

1. a3= 24

a6= 192

a+2d= 24

a+5d=192(sub)

-3d= -168

d=-168/-3

d = 56

a+2(56)= 24

a+ 112=24

a= 24-112

a= -88

a10 = a+9d

a10 = -88+9(56)

= -88+504

= 416

2. 6,12,18........(40 terms)

a= 6

d= 12-6= 6

n= 40

Sn= n/2[2a+(n-1)d]

S40= 40/2[2(6)+(40-1)6]

= 20 [12+39(6)]

= 20[12+234]

= 20(246)

= 4920

3. 5,11,17,23.........301

a= 5

d= 11 -5

= 6

an= a+(n-1)d

301= 5+(n-1)6

301= 5+6n-6

301 = 6n-1

301+1= 6n

302 = 6n

n= 302/6

n= 50.33

301 is not a term in the list of the numbers

Answer 2

Answer:

4). 3rd term = ar^2 = 24. & ar^5 = 192

So, just divide them,

ar^5/ar^2 = 192/24

r^3 = 8

So, r = 2

Now, 3rd term = ar^2 = 24

a = 24/4 = 6

So, 10th term = ar^9 = 6(2)^9 = 3072_____ (Ans.)

5). Sum of first 40 positive integers divisible by 6 is in AP

Where a = 6 ; d = 6 & n = 40

Sn = n/2(2a+(n-1)d)

On calculating

We get Sn = 4920______(Ans.)

6). No 301 is not the term of the given sequence.

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