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Formula for sum of nn terms: Sn=n2(2a+(n−1)d)S_n = \frac{n}{2}(2a + (n-1)d). Given S20=710S_{20}=710, a=5a=5, n=20n=20.

710=202(2⋅5+19d)=10(10+19d).710 = \frac{20}{2}(2\cdot 5 + 19d) = 10(10 + 19d).

Divide both sides: 71=10+19d⇒19d=61⇒d=61/19=3.2171 = 10 + 19d \Rightarrow 19d = 61 \Rightarrow d = 61/19 = 3.21. Wait, check arithmetic: 710/10=71710/10 = 71. Yes. Then 71=10+19d⇒61=19d⇒d=61/19=3.2171 = 10 + 19d\Rightarrow 61=19d\Rightarrow d=61/19=3.21. That is not an integer. Re-check setup. Formula correct. Let’s try alternative: Average of first and last term × number of terms. Let last term = a+19da+19d. Then sum = (20/2)(first+last) = 10(5 + (5+19d)) = 10(10+19d). Same as before. So correct. Compute again: 710/10 = 71. 71–10=61. 61/19=3.21, not neat. But 61/19=3.2105, maybe answer choices? They gave 2,3,4,5. Closest is 3. Did I mis-copy? Let’s recompute carefully. If d=3, last term=5+193=62. Sum = 20(5+62)/2=2067/2=670. Too small. If d=4, last=81. Sum=20(86)/2=860. Too large. So actual is between 3 and 4. Correct d=61/19=3.21. Maybe test intends 3.2 ~ 3.21 → not listed properly. The mathematically correct common difference is 61/19 ≈ 3.21.