Question

16. Find the sum of all two digit odd positive numbers​

Answer 1

Step-by-step explanation:

the all two odd positive number are

11, 13, 15, 17-----------99

according to n term

so,

a=11

d=13-11=2

TN=99

now from formula

Tn=a+(n-1)d

99=11+(n-1)2

88=(n-1)2

n-1=44

:-n=44+1=45

now sum of all n term by formula of Sn

so,

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

=45/2{(2*11+(45-1)2}

=45/2(22+88)

=45/2*88

=45*44

=1980

hence sum of all two odd numbers will be 1980

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Answer 2

$\huge \pink { \boxed{ \boxed{ \mathsf{ \mid \ulcorner Answer :- \urcorner \mid }}}}$

★ Given :-

A.P :- 11, 13, 15, 17, 19, 21 ..................99

First term (a) = 11

Common difference (d) = 13 - 11 = 2

last term (an/L) = 99

★ To Find :-

Sum of all two digit odd numbers.

★ Solution :-

We know that,

$\huge{\boxed{\blue{\sf{a_{n} = a + (n - 1)d}}}}$

______________[Put Values]

99 = 11 + (n - 1)2

99 = 11 + 2n - 2

99 = 9 + 2n

99 - 9 = 2n

90 = 2n

90 / 2 = n

45 = n

$\huge{\boxed{\red{\sf{n = 45}}}}$

$\rule{200}{2}$

Now, we know that

$\huge{\boxed{\blue{\sf{S_{n} = \frac{n}{2}(a + L)}}}}$

______________[Put Values]

Sn = 45 * (11 + 99) / 2

Sn = (45 * 110) / 2

Sn = 4950 / 2

Sn = 2475

$\huge{\boxed{\red{\sf{S_{n} = 2475}}}}$