Question

In an A.P. of 50 terms, the sum of first 10 terms is 210 and the sum of its

last 15 terms is 2565. Find the A.P.​

Answer 1

Answer: It's very easy ,

You've to just find value of d or a , and put in your equation.

Step-by-step explanation:

1. S10 = 10/2 (2a +9d) = 210

= 2a +9d = 210/5

= 42

So, 2a = 42-9d

Now,

Acc. to Q,

Sum of last 15 terms

= S50 - S35 = 2565

= 50/2 (2a+49d) - 35/2 (2a+34d) = 2565.

25(2a+49d) - 35/2(2a+34d) =2565

Put 2a = 42-9d, in this eqn.

We get,

25(42-9d+49d) - 35/2(42-9d+34d) = 2565

= 25(42+40d) - 35/2(42 +25d)= 2565.

= 1050+1000d - 735 - 875/2 d = 2565.

= 2000d-875d/2 = 2565-315

= 1125d = 2250 x 2

So,, d = 2250x2/1125 = 4.

Hurray! We got d, now out d=4 , in any eq.

Here, from topmost eq,

5(2a+9d) = 210

10a = 210- 9x4

A= 30/10 = 3.

So, ap is 3,7,11,15,19,23......

Mark this as brainlist

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{ A.P=3,7,11,15...}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explantion:}}}}$

• In the given question information given about an A. P. of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565.• We have to find the A.P.

$\green{\underline \bold{Given : }} \\ :\implies \text{Number \: of \: terms(n) = 50 }\\ \\ :\implies S_{10} = 210 \\ \\ :\implies S_{15}(From \: last) = 2565 \\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies A.P = ?$

• According to given question :

$\bold{For \: sum \: of \: 10th \: term : } \\ :\implies S_{10} = \frac{10}{2} (2a + (10 - 1) \times d)\\ \\ :\implies 210 = 5(2a + 9d) \\ \\ :\implies \frac{210}{5} = 2a + 9d \\ \\ :\implies 2a + 9d = 42 - - - - - (1) \\ \\ \bold{For \: sum \: of \: last \: 15 \: terms : } \\ \\ :\implies S_{15}(from \: end) = \frac{15}{2} (2 \times a_{36} + (15 - 1) \times d) \\ \\ :\implies 2565 = \frac{15}{2} (2 \times (a + 35d) + 14d)\\ \\ :\implies 2565 = \frac{15}{2} (2a + 70d + 14d) \\ \\ :\implies 2565 \times 2 = 15(2a + 84d) \\ \\ :\implies \frac{ \cancel{2565} \times 2} { \cancel{15}} = 2(a + 42d) \\ \\ :\implies a + 42d = \frac{171 \times \cancel2}{ \cancel2}$

$\\ :\implies a + 42d = 171 \\ \\ :\implies a = 171 - 42d - - - - - (2) \\ \\ \bold{Putting \: value \: of \: a \: in \: (1)} \\ :\implies 2(171 - 42d) + 9d = 42 \\ \\ :\implies 342 - 84d + 9d = 42 \\ \\ \implies - 75d=42 - 342 \\ \\ :\implies \cancel - 75d = \cancel- 300 \\ \\ :\implies d = \frac{300}{75} \\ \\ \bold{:\implies d = 4} \\ \\ \bold{Putting \: value \: of \: d \: in \:(1)} \\ :\implies 2a + 9 \times 4 = 42 \\ \\ :\implies 2a + 36 = 42 \\ \\ :\implies 2a + 450 = 484 \\ \\ :\implies 2a = 42 - 36 \\ \\ :\implies a = \frac{6}{2} \\ \\ \bold{:\implies a =3} \\ \\ \green{\therefore {\text{A.P = 3,7,11,15....}}}$