Question

The th n term of an A.P. is 3n −1 .Choose from the following the sum of its first five terms

Answer 1

Answer:

Step-by-step explanation:

nth term is 3n-1

for 1st term, n=1

3×1-1 ( putting the value of n)

3-1

2

Therefore 1st term = 2

for 2nd term, n=2

3×2-1 ( putting the value of n)

6-1

5

Therefore the 2nd term is 5

Now, d = 2nd term - 1st term

d = 5 - 2

d = 3

Now, n = 5 ( since we need to find the first 5 terms)

Now the formula for sum -----

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

Putting the values in the formula ----

AP = 5/2 [ 2×2 + ( 5-1)×3 ]

AP = 5/2 [ 4 + 4×3 ]

AP = 5/2 [ 4 + 12 ]

AP = 5/2 × 16

AP = 5×8

AP = 40

Therefore, 40 is the answer.

Answer 2

Question  :-

     Q) The th n term of an A.P. is 3n −1 .Choose from the following the sum of its first five terms

Answer :-

$a_{n} = 3n-1$

 ∴ $3n-1 = a + (n-1)d$

∴ $a_{1} = 3n-1 = 3(1) - 1 =2$

  $a_{2} = 3n-1 = 3(2)-1 =5$

  $a_{3} = 3n-1 = 3(3) -1 = 9-1 =8$

∴ $d = a_{2} - a_{1} = 5-2 = 3$

     $= a_{3} - a_{2} = 8-5 = 3$

as , a = 2

∴ sum of $s_{n}$

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

$s_{n} = \frac{5}{2} [ 2*2 + (5-1)*3]$

$s_{n} = \frac{5}{4} [ 4+12]$

$s_{n} = 5*16/2$

$s_{n} = 40$

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