Question

[Class 10, Arithmetic Progression]
Q. Two APs have the same common difference. If the first term of these APs be 3 and 8 respectively, find the difference between the sums of their 50 terms. ​

Answer 1

Answer:

250

Step-by-step explanation:

For first AP

First term, a = 3

Commom Difference, = d

We know that,

Sum of n terms of an AP is given by,

• Sn = n/2{2a+(n-1)d}

Therefore, we get,

=> S_50 = 50/2{2×3 + (50-1)d}

=> S_50 = 25 (6 +49d) ..........(1)

Now, for second AP.

First term = 8

Common Difference = d

Therefore, we have,

=> s_50 = 50/2{2×8 + (50-1)d}

=> s_50 = 25 (16 + 49d) ........(2)

Subtracting eqn (1) from (2), we get,

=> D = 25{(16+49d) - (6 + 49d)}

=> D = 25{(16-6) + (49-49)d}

=> D = 25{10}

=> D = 250

Hence, the required difference is 250.

Answer 2

Answer:

Let Common Difference of Both APs be d.

• 1st Term of First AP (a₁) = 3
• 1st Term of Second AP (a₂) = 8

☢ $\underline{\textsf{Difference between the sum of 50 terms :}}$

$:\implies\sf S_{50}\:of\:2nd\:AP-S_{50}\:of\:1st\:AP\\\\\\:\implies\sf \bigg\lgroup\dfrac{n}{2}\bigg(2a_2-(n-1)d\bigg)\bigg\rgroup-\bigg\lgroup\dfrac{n}{2}\bigg(2a_1-(n-1)d\bigg)\bigg\rgroup\\\\\\:\implies\sf \bigg\lgroup\dfrac{50}{2}\bigg(2 \times 8-(50-1)d\bigg)\bigg\rgroup-\bigg\lgroup\dfrac{50}{2}\bigg(2 \times 3-(50-1)d\bigg)\bigg\rgroup\\\\\\:\implies\sf \bigg\lgroup25(16 - 49d)\bigg\rgroup - \bigg\lgroup25(6 - 49d)\bigg\rgroup\\\\\\:\implies\sf \bigg\lgroup(25 \times 16) - (25 \times 49d)\bigg\rgroup - \bigg\lgroup(25 \times 6) - (25 \times 49d)\bigg\rgroup\\\\\\:\implies\sf (25 \times 16) - \bcancel{(25 \times 49d)} - (25 \times 6) + \bcancel{(25 \times 49d)}\\\\\\:\implies\sf (25 \times 16) - (25 \times 6)\\\\\\:\implies\sf 25(16 - 6)\\\\\\:\implies\sf 25 \times 10\\\\\\:\implies\underline{\boxed{\sf 250}}$

⠀

$\therefore\:\underline{\textsf{Difference b/w their sum of 50 terms is \textbf{250}}}.$