Question

in an ap of 50 terms, the sum of first 10 erms is 210 and the sum of its last 15 terms is 2565. Find the A.P.

Answer 1

Answer:

Step-by-step explanation:

n=50

a₁₀=a+9d

a=-9d

s₁₀=10/2[a+a+9d]

    =5[2a+9d]

    =10a+45d=210

    =2a+9d=42 ...............(i)

a₃₆=a+35d

s₁₅₍from last₎=15/2[a+35d+a+49d]

                    =15/2[2a+84d]

                    =15a+630d=2565

                    =a+42d=171.......................(ii)

solving (i) and(ii)

2a+9d=42

a+42d=171

⇒d=4

a=39

∴ap=39,43,47,51,55,59,63............

   

Answer 2

Let the A.P be : a , a + d , a + 2d. . . . . . .a + 35d. . . . . . .a + 49d

Given : The Sum of First 10 Terms of the A.P is 210

$\mathsf{We\;know\;that,\;Sum\;of\;(n)\;terms\;in\;an\;A.P\;is\;given\;by :}$

$\mathsf{\implies S_n = \frac{n}{2}[2a + (n - 1)d]}$

$\mathsf{\implies Sum\;of\;First\;10\;terms\;(S_1_0) = \frac{10}{2}[2a + (10 - 1)d]}$

$\mathsf{\implies 5[2a + 9d] = 210}$

$\mathsf{\implies [2a + 9d] = 42\;----------\;(1)}$

Given : The Sum of Last 15 terms of the A.P is 2565

We need to Realize that : In the Series of Last 15 Terms, The First Term will be 36th term which can be written as (a + 35d) and The Common Difference Remains the Same.

$\mathsf{\implies Sum\;of\;the\;Last\;15\;terms\;(S_1_5_L) = \frac{15}{2}[2(a + 35d) + (15 - 1)d]}$

$\mathsf{\implies \frac{15}{2}[2(a + 35d) + 14d] = 2565}$

$\mathsf{\implies \frac{1}{2}[2(a + 35d) + 14d] = 171}$

$\mathsf{\implies [2(a + 35d) + 14d] = 342}$

$\mathsf{\implies [2a + 84d] = 342\;----------\;(2)}$

Subtracting Equation (1) from Equation (2), We get :

⇒ [2a + 84d] - [2a + 9d] = 342 - 42

⇒ 75d = 300

⇒ d = 4

Substituting d = 4 in Equation (1), We get :

⇒ 2a + 9(4) = 42

⇒ 2a + 36 = 42

⇒ 2a = 6

⇒ a = 3

Hence, The Given Arithmetic Progression is :

✿  3 , 7 , 11 , 15. . . . . . . . . .143. . . . . . . . . .199