Question

difference of first terms of two arithmetic sequences with same common difference is 10

what is the difference between second term of these sequence ​

Answer 1

Answer:-

Let the first term of the 1st AP be "a1" and the 2nd AP be "a2".

Let the common differences of two APs be "d".

Given:

Difference of their 1st terms = 10

→ a1 - a2 = 10 -- equation (1)

We know that,

nth term of an AP = a + (n - 1)d.

Hence,

Second term of first AP a(2) = a1 + (2 - 1)d

→ a(2) = a1 + d

Second term of second AP a(2) = a2 + (2 - 1)d

→ a(2) = a2 + d

Difference of their 2nd terms = a(2) of first AP - a(2) of second AP.

→ Difference of their 2nd terms = a1 + d - (a2 + d)

→ Difference of their 2nd terms = a1 + d - a2 - d

→ Difference of their 2nd terms = a1 - a2

Putting the value of "a1 - a2" from equation (1) we get,

→ Difference of their 2nd terms = 10.

Therefore, the difference between the second terms of two APs is also 10.

Answer 2

$\large\mathcal{\underbrace{SOLUTION:-}}$

✍️ Let,

• First Arithmetic progression = $\rm{A.P_1}$

• Second Arithmetic progression = $\rm{A.P_2}$

• First term of $\rm{A.P_1}$ = x

• First term of $\rm{A.P_2}$ = y

✍️ Given that common difference is same for both arithmetic sequence .

• So, Common Difference = d

$\checkmark\:\mathcal{\purple{\underline{n^{th}\:term\:of\:an\:A.P\:=\:a\:+\:(n\:-\:1)\:d}}}$

✍️ According to the question,

• x - y = 10 ------(1)

✍️ Second term of $\rm{A.P_1}$,

$\rm{\implies\:{(a_2)}_1\:=\:x\:+\:(2\:-\:1)\:d}$

$\rm{\red{\implies\:{(a_2)}_1\:=\:x\:+\:d}}$

✍️ And second term of $\rm{A.P_2}$

$\rm{\implies\:{(a_2)}_2\:=\:y\:+\:(2\:-\:1)\:d}$

$\rm{\red{\implies\:{(a_2)}_2\:=\:y\:+\:d}}$

✍️ Now, difference between second term of $\rm{A.P_1\:and\:A.P_2}$ is,

$\rm{=\:x\:+\:d\:-\:(y\:+\:d)\:}$

$\rm{=\:x\:+\:d\:-\:y\:-\:d\:}$

$\rm{=\:x\:-\:y\:}$

✔️ put the value of ‘x - y = 10’ in the above equation, we get

$\huge\rm{\blue{=\:10\:}}$

$\bigstar\:\mathcal{\green{\:Difference\:between\:2nd\:term\:of\:A.P_1\:and\:A.P_2\:=\:10\:}}$