Question

the sum of the first seven terms of an arithmetic progression is 140 and the sum of the next 7 terms of the same progression is 385 then find the arithmetic progression​

Answer 1

Step-by-step explanation:

Given that the sum of first 7 terms is 140.

We can say that:-

$\sf{S_7 = \dfrac{7}{2}(2a + (7-1) \times d)}$

$\sf{140= \dfrac{7}{2}(2a + 6d)}$

$\sf{\dfrac{140 \times 2}{7}= 2a + 6d}$

$\sf{20 \times 2 = 2a + 6d}$

$\sf{ 40 = 2a + 6d}$

$\sf{20 = a + 3d}$

$\sf{20 - 3d = a }$ -------------(1)

Now, sum of the next 7 terms of the same progression is 385. So, we can say that the sum is 14.

$\sf{S_{14} = \dfrac{14}{2}(2a + (14-1) \times d)}$

Now, 140 + 385 = 525

$\sf{525 = 7(2a + 13d)}$

$\sf{\dfrac{525}{7} = 2a + 13d}$

$\sf{75= 2a + 13d}$

Substitute the value of (1) here,

$\sf{75= 2(20 - 3d) + 13d}$

$\sf{75= 40 - 6d + 13d}$

$\sf{75= 40 + 7d}$

$\sf{75 - 40= 7d}$

$\sf{ 35= 7d}$

$\sf{ 5 = d}$

Finding the a of the term,

$\sf{20 - 3d = a }$

$\sf{20 - 15= a }$

$\sf{ 5 = a }$

$\sf{a_n = 5 + 3 (5)}$ = 75.

So, the sequence will be 5, 10, 15 ... till 75.

Answer 2

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• The sum of the first seven terms of an arithmetic progression is 140 and the sum of the next 7 terms of the same progression is 385.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• The Arithmetic progression

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

According to the question,

• The sum of the first 7 terms of an arithmetic progression is 140.

We know that,

$\boxed{\pink{\sf S_n \: = \: \dfrac{n}{2} (2a \: + \: (n \: - \: 1) d)}}$

• Here,

• n = 7

• $\sf S_7$ = 140

Subsituting the values,

• $\sf 140 \: = \: \dfrac{7}{2} (2a \: + \: (7 \: - \: 1) d)$

• $\sf 140 \: = \: \dfrac{7}{2} (2a \: + \: 6d)$

• $\sf 140 \: * \: \dfrac{2}{7} \: = \:2a \: + \: 6d$

• $\sf \cancel{140} \: * \: \dfrac{2}{\cancel{7}} \: = \:2a \: + \: 6d$

• $\sf 20 \: * \: \dfrac{2}{1} \: = \: 2a \: + \: 6d$

• $\sf 20 \: * \: 2 \: = \: 2a \: + \: 6d$

• $\sf 40 \: = \: 2a \: + \: 6d$

• $\sf 40 \: = \: 2(a \: + \: 3d)$

• $\sf \dfrac{40}{2} \: = \: a \: + \: 3d$

• $\sf \dfrac{\cancel{40}}{\cancel{2}} \: = \: a \: + \: 3d$

• $\sf 20 \: = \: a \: + 3d$

• $\sf 20 \: - \: 3d \: = \: a \: \longrightarrow \boxed{1}$

Now,

• The sum of the next 7 terms of the same progression is 385.

We know that,

$\boxed{\pink{\sf S_n \: = \: \dfrac{n}{2} (2a \: + \: (n \: - \: 1) d)}}$

• Here,

• n = 14

• Here, it is asked as the next progression,

• So, the sum of first 7 terms progression and the sum of next 7 terms progression must be added.

• $\sf S_{14}$ = 140 + 385

• $\sf S_{14}$ = 525

Subsituting the values,

• $\sf 525 \: = \: \dfrac{14}{2} (2a \: + \: (14 \: - \: 1) d)$

• $\sf 525 \: = \: \dfrac{\cancel{14}}{\cancel{2}} (2a \: + \: 13d)$

• $\sf 525 \: = \: 7 (2a \: + \: 13d)$

• $\sf \dfrac{525}{7} \: = \: 2a \: + \: 13d$

• $\sf \dfrac{\cancel{525}}{\cancel{7}} \: = \: 2a \: + \: 13d$

• $\sf 75 \: = \: 2a \: + \: 13d$

Substituting $\boxed{1}$ in the above equation,

• $\sf 75 \: = \: 2a \: + \: 13d$

• $\sf 75 \: = \: 2(20 \: - \: 3d) \: + \: 13d$

• $\sf 75 \: = \: 40 \: - \: 6d \: + \: 13d$

• $\sf 75 \: - \: 40 \: = \: 13d \: - \: 6d$

• $\sf 35 \: = \: 13d \: - \: 6d$

• $\sf 35 \: = \: 7d$

• $\sf \dfrac{35}{7} \: = \: d$

• $\sf \dfrac{\cancel{35}}{\cancel{7}} \: = \: d$

Therefore,

• d = 5

Substituting d = 5 in $\boxed{1}$ ,

• 20 - 3d = a

• 20 - 3 * 5 = a

• 20 - 15 = a

• 5 = a

Therefore,

• a = 5

Now,

• Arithmetic progression =

a = 5

$\sf a_2$ = a + d = 5 + 5 = 10

$\sf a_3$ = a + 2d = 5 + 2 * 5 = 5 + 10 = 15

$\sf a_4$ = a + 3d = 5 + 3 * 5 = 5 + 15 = 20

Therefore,

• $\underline{\boxed{\therefore{\purple{\sf The \: Arithmetic \: Progression \: = \: 5, \: 10, \: 15, \: 20, \: ...}}}}$