This question's explanation and your answers so far.
流れ図Xで示す処理では，変数 i の値が，1→3→7→13と変化し，流れ図Yで示す処理では，変数 i の値が，1→5→13→25と変化した。図中の a，b に入れる字句の適切な組合せはどれか。 〔流れ図X〕 開始 → i に1を格納する → 変換（ループ端） k：1，1，3 （注） → ［ a ］を計算し，その結果を i に格納する → 変換（ループ端） → 終了 〔流れ図Y〕 開始 → i に1を格納する → 変換（ループ端） ［ b ］ （注） → ［ a ］を計算し，その結果を i に格納する → 変換（ループ端） → 終了 （注）ループ端の繰返し指定は，変数名：初期値，増分，終値を示す。
a=2i+k is wrong. Checking in flow diagram X: the 1st pass gives 2×1+1=3, which matches, but the 2nd pass gives 2×3+2=8, which disagrees with the actual value 7. Since the expression for a already fails to hold, this option is inappropriate. b=k:1,3,7 is also wrong, as described below.
b=k:2,2,6 is correct, but a=2i+k is wrong. As with option ア, the 2nd pass in flow diagram X gives 2×3+2=8 instead of 7. Since a does not match, the option as a whole is inappropriate.
a=i+2k is correct, but b=k:1,3,7 is wrong. It specifies looping k with initial value 1, increment 3, final value 7, that is, k=1,4,7. Substituting into the 1st pass of flow diagram Y gives 1+2×1=3, not the actual value 5. Since Y cannot be reproduced, this option is inappropriate.
正解 First determine a from flow diagram X. In X, the initial value of i is 1, and the loop terminal specifies k:1,1,3 (k has initial value 1, increment 1, final value 3), so it loops 3 times with k=1,2,3. Substituting a=i+2k on each pass: 1st pass 1+2×1=3, 2nd pass 3+2×2=7, 3rd pass 7+2×3=13, which exactly matches the change of i, namely 1→3→7→13. Therefore a=i+2k. Next, determine b from flow diagram Y. Y uses the same a=i+2k, and i changes as 1→5→13→25. Working backward to find the k needed for each transition: for 1→5, from 5=1+2k we get k=2; for 5→13, k=4; for 13→25, k=6. Thus k increases from 2 by steps of 2 up to 6, that is, b=k:2,2,6. The only option satisfying both a=i+2k and b=k:2,2,6 is エ.