Question

Two arithmetic progression has the same common difference . If the first term of the first progression is 3 and that of the other is 8,then the difference between their third term is​

Answer 1

Given: Two Arithmetic Progression has the Same Common Difference ( d ). & the First term of the First AP is 3 and First term of Other AP is 8.

Need to find: The Difference b/w their 3rd term?

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As we know that,

To Calculate the nth term of an AP(Arithmetic Progression) the formula is Given by :${\underline{\pmb{\sf{T_n = a + (n - 1)d}}}}$

where:

• a = First Term
• n = no. of Terms
• d = Common difference

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Therefore,

⟩⟩ Third term of 1st AP;

• Common difference, (d) = d
• First Term = 3

$:\implies\sf T_3 = 3 + 2d\qquad\quad\sf\Bigg\lgroup eq^{n}\;(1)\Bigg\rgroup$

⟩⟩ Third term of 2nd AP;

• Common difference, (d) = d
• First Term = 8

$:\implies\sf T_3 = 8 + 2d\qquad\quad\sf\Bigg\lgroup eq^{n}\;(2)\Bigg\rgroup$

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✇ Now, With the help of both eqₙ ( 1 ) & eqₙ ( 2 ) finding the distance b/w the 3rd term of AP —

$:\implies\sf \Big(3 + 2d \Big) - \Big(8 + 2d \Big) \\\\:\implies{\pmb{\frak{- \: 5}}}$

∴ Hence, the Difference b/w their third Term is – 5.

Answer 2

Given :-

Two arithmetic progression has the same common difference . If the first term of the first progression is 3 and that of the other is 8

To Find :-

Difference between their third term is​

Solution :-

We know that

$\sf S_{n}=a+(n-1)d$

◼ I n C a s e 1

We have

a = first term = 3

d = common difference = d

By putting the values

$\sf S_{3} = 3+(3-1)d$

$\sf S_{3}=3+2d$

◼ I n C a s e 2

We have

a = first term = 8

d = common difference = d

By putting the values

$\sf S'_{3} = 8+(3-1)d$

$\sf S'_{3}=8+2d$

Now

$\sf 3^{rd}\;term = S_{3}-S'_{3}$

$\sf 3^{rd}\;term=3+2d-(8+2d)$

$\sf 3^{rd}\;term=3+2d-8-2d$

$\sf 3^{rd}\;term=3-8$

$\sf 3^{rd}\;term=-5$