Question

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

Answer 1

Answer:

• Hope the concept is clear!

Step-by-step explanation:

${\underline{\underline{\star{\sf{Given:}}}}}\\\\\bf{S_7(Sum\:of\:first\:7\:terms) = 140}\\\\ \bf{Sum\:of\:next\:7\:terms = 385}$

So, Sum of the first 14 terms = Sum of the first seven terms + Sum of the next 7 terms.

• $\bf{S_{14} = 140+385}$
• $\bf{S_{14} = 525}$

${\underline{\underline{\star{\sf{Concept:}}}}}$

Here the concept of finding the sum of $\tt{n_{th}}$ term is used. In this question, first we are going to arrive an equation of Sum of 7 terms. Second, we are going to arrive an equation of Sum of 14 terms. Then we are going to substitute the equations to get the value of a and d. Then, at last, we will find the A.P

${\underline{\underline{\star{\sf{Formula\:used:}}}}}$

$\underline{\boxed{\orange{\sf{S_n = \dfrac{n}{2}[2a+(n-1)d]}}}}$

${\underline{\underline{\star{\sf{Solution:}}}}}$

$\implies\sf{S_7 = \dfrac{7}{2}[2a+(7-1)d]}\\\\\implies\sf{140 = \dfrac{7}{2}[2a+6d]}\\\\\implies\sf{2a+6d = \dfrac{140 \times 2}{7}}\\\\\implies\sf{2a+6d = 40}\\\\\implies\sf{2a = 40 - 6d}\\\\\implies\sf{a = \dfrac{40-6d}{2}}\\\\\implies\boxed{\pink{\tt{a = 20 - 3d}}}$

Similarly,

$\implies\sf{S_{14} = \dfrac{14}{2}[2a+(14-1)d]}\\\\\implies\sf{525 = 7[2a+13d]}\\\\\implies\sf{2a+13d= \dfrac{525}{7}}\\\\\implies{\boxed{\pink{\tt{2a+13d = 75}}}}$

Now, substitute the value of $\tt{a = 20 - 3d}$ in $\tt{2a+13d = 75}$

$\implies\sf{2(20-3d)+13d = 75}\\\\\implies\sf{40-6d+13d = 75}\\\\\implies\sf{40+7d = 75}\\\\\implies\sf{7d = 35}\\\\\implies{\boxed{\tt{d = 5}}}$

Substituting the value of $\tt{d=5}$ in $\tt{a = 20 - 3d}$

$\implies\sf{a = 20 -3 \times 5}\\\\\implies\sf{a = 20 -15}\\\\\implies{\boxed{\tt{a =5}}}$

So, we found:

• a = 5
• d = 5

So, the A.P will be: 5, (5+5), (2×5+5), (3×5+5)

• 5,10,15,20...

Answer 2

this answer is correct

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