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Arithmetic Progressions — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsArithmetic Progressions3 Marks
Question
Prove that no matter what the real numbers $a$ and $b$ are, the sequence with $n^{th}$ term $a + nb$ is always an $A.P.$ What is the common difference$?$
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Answer

$ \text { Given sequence }\left(a_n\right)=a_n+6 n$
$n^{\text {th }} \text { term }\left(a_n\right)=a+n b$
$(n+1)^{\text {th }} \text { term }\left(a_{n+1}\right)=a+(n+1) b$
$\text { Common difference }(d)=a_{n+1}-a_n$
$d=(a+(n+1) b)-(a+n b)$
$=a+n b+b-a-n b$
$=b$
$\therefore$ Common difference $(d)$ does not depend on $n^{th}$ value so, given sequence is in $AP$ with $(d) = b.$

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