Question

Five terms are inserted between 1 and 100 so that the resulting sequence forms an AP. Find the common difference of the AP.

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Answer 1

Solution:-

We have to intserted five term between 1 and 100

we can write as

$\rm \to1, \: a_1, \: a_2, \: a_3, \: a_4, \: a_5, \: 100$

Now we get

$\rm \: first \: term \: (a) = 1$

$\rm \: a_7 = 100$

Using formula

$\boxed{ \rm \: T_n =a + (n - 1)d}$

$\rm \: 100 = 1 + (7 - 1)d$

$\rm \: 100 - 1 = 6d$

$\rm99 = 6d$

$\rm \: d = \dfrac{33}{2}$

So common differences is 33/2

Answer 2

Step-by-step explanation:

We have to insert 5 terms between 1 and 100

we can write as

a1,a2,a3 a5,a6 ,100

first term (a)=1

Tn=a+(n-1)d

100=1+(7-1) d

100-1=6d

99=6d

d=99/6

d=33/2