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Exercise 5.5 · Q10

Q.If SnS_n denotes the sum of nn terms of an AP whose common difference is dd, the value of Sn−2Sn−1+Sn−2S_n-2S_{n-1}+S_{n-2} is

(1) 00
(2) 2d2d
(3) 4d4d
(4) d2d^2
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Recognise Sn−2Sn−1+Sn−2S_n-2S_{n-1}+S_{n-2} as (Sn−Sn−1)−(Sn−1−Sn−2)=Tn−Tn−1(S_n-S_{n-1})-(S_{n-1}-S_{n-2})=T_n-T_{n-1}, which is simply the common difference dd — verified independently by direct substitution of the SnS_n formula.

Step 1. Regroup as a difference of consecutive terms.

Sn−2Sn−1+Sn−2=(Sn−Sn−1)−(Sn−1−Sn−2)=Tn−Tn−1S_n-2S_{n-1}+S_{n-2} = (S_n-S_{n-1})-(S_{n-1}-S_{n-2}) = T_n - T_{n-1}

(since Sn−Sn−1S_n-S_{n-1} is simply the nthn^{th} term TnT_n, and Sn−1−Sn−2S_{n-1}-S_{n-2} is the (n−1)th(n-1)^{th} term Tn−1T_{n-1}).

Step 2. Use the AP definition. Tn−Tn−1=dT_n-T_{n-1}=d, the common difference, for EVERY AP.

Step 3. Verify directly with the closed-form Sn=na+n(n−1)2dS_n=na+\frac{n(n-1)}2d. Substituting and expanding Sn−2Sn−1+Sn−2S_n-2S_{n-1}+S_{n-2} fully in terms of a,d,na,d,n, every aa-term and every n2,nn^2,n-term cancels, leaving exactly the constant dd.

Step 4. Numerical check. AP 1,4,7,10,13,…1,4,7,10,13,\ldots (a=1,d=3a=1,d=3): S2=5,S3=12,S4=22S_2=5,S_3=12,S_4=22. S4−2S3+S2=22−24+5=3=dS_4-2S_3+S_2=22-24+5=3=d ✓ (not 2d=62d=6, 4d=124d=12, or d2=9d^2=9). …

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