Question

If the sum of n terms of an AP is Sn = 3n2 + 5 then a100 -a99 =6
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Answer 1

The value of  $a_1_0_0$  -  $a_9_9$  is 1 .

Step-by-step explanation:

Given as :

For Arithmetic Progression

Sum of n terms of an A.P = 3 n² + 5

∵ $S_n$  = $\dfrac{n}{2}$ [ 2 a + ( n - 1 ) d ]

where   a is the first term

             d is the difference

For n = 1

∵ $S_n$  =  3 n² + 5

  $S_1$     = 3 ×1² + 5

  $S_1$     = 8

So, First term = a = 8

For n = 2

$S_n$  = 3 × 2² + 5

 $S_2$  = 12 + 5

  $S_2$ = 17

So, Sum of Second term = 17

Again

Sum of second term = First term + second term

∴ second term = Sum of second term - First term

                       = $S_2$ - a

                       = 17 - 8  = 9

 So, Second term = 9

∴  Common difference = Second term - first term

                                  d = 9 - 8

Or,                               d = 1

So, The common difference = d = 1

Again ,

∵   nth term = $a_n$  = a + ( n -1 ) d

Now, For n = 100 term

        $a_1_0_0$ = a + (100 - 1)d

               = a + 99 × d

               = a + 99 d

∴    $a_1_0_0$ =   a + 99 d

And For n = 99 term

        $a_9_9$ = a + (99 - 1)d

               = a + 98 × d

  ∴  $a_9_9$  = a + 98 × d

Now,

The value of     $a_1_0_0$  -  $a_9_9$  = (a + 99 d)  - ( a + 98 d)

                                           = ( a - a) + (99 d - 98 d)

                                            = 0 + d

∴  $a_1_0_0$  -  $a_9_9$  = d = 1

Hence, The value of  $a_1_0_0$  -  $a_9_9$  is 1 . Answer

Answer 2

$a_{100} -a_{99 }$ = 1

Step-by-step explanation:

Given,

$S_n$ = 3$n^2$ + 5

To find, $a_{100} -a_{99 }$ = ?

Put n = 1, 2, 3, 4, 5, ........

$S_1$ = 3$(1)^2$ + 5 = 3 + 5 = 8

$S_1$ = First term (a) = 8

$S_2$ = 3$(2)^2$ + 5 = 12 + 5 = 17

$S_2$ =  First term + Second term = 17

⇒ Second term = 17 - 8 = 9

∴ Common difference = Second term - First term

= 9 - 8

= 1

∴ $a_{100}$ = a + 99d and

$a_{99}$ = a + 98d

∴ $a_{100} -a_{99 }$ = a + 99d  - (a + 98d)

= a + 99d  - a - 98d

= d

= 1

∴ $a_{100} -a_{99 }$ = 1