Question

The sum of all 2 -digit odd numbers is :
(A) 2475
(C) 4905
(B) 2530
(D) 5049​

Answer 1

Answer:

(C)4905

Step-by-step explanation:

Sn = 90/2 [2(10)+(90-1)1] Sn = 45 [20+89] Sn = 45[109] Sn = 4905 therefore, sum of all 2 digit numbers = 4905 2nd method: Sn = n/2 ...

Answer 2

Answer:

$\huge{\green{\boxed{\sf{(A)2475}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

▪In the given question we have information about a series of numbers whose difference is equal.

▪So we have to find the sum of that series.

• A.P=11,13,15,..........,99

$\: \: \underline\bold{given : } \\ \to a = 11\\ \to c.d = 2 \\ \to a_{n} = 99 \\ \\ \: \underline\bold{to \: find : } \\ \to sum \: of \: two \: term \: odd \: number = ?$

▪According to given question :

$\to a_{n} = a + (n - 1)d \\ \to 99 = 11 + (n - 1) \times 2 \\ \to 99 - 11 = (n - 1) \times 2 \\ \to \frac{88}{2} = n - 1 \\ \to n = 44 + 1 \\ \bold{\to n = 45}$

▪We find how many terms in this A.P so we put those value in formula of sum of nth terms.

$\to s_{n} = \frac{n}{2} (a + a_{n}) \\ \to s_{45} = \frac{45}{2} (11 + 99) \\ \to s_{45} = \frac{45}{2} \times 110 \\ \to s_{45} = 45 \times 55 \\ \bold{\to s_{45} = 2475}$

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